A block slides on a rough horizontal surface from point A to point B. A force (magnitude P = 1) -3.3 J
2.0 N) acts on the block between A and B, as shown. Points A and B are 1.5 m apart. If the
kinetic energies of the block at A and B are 5.0 J and 4.0 J, respectively, how much work is Wf = (Kef-Kei)-P*costheta*d =
done on the block by the force of friction as the block moves from A to B? (4-5) - 2cos40(1.5) = -3.3 J
1) -3.3 J Not on formula sheet
2) +1.3 J
3) +3.3 J
4) -1.3 J
5) +4.6 J
A constant force of 15 N in the negative y direction acts on a particle as it moves from the origin
to the point (3i+3j-k) m. How much work is done by the given force during this displacement? 2) -45 J
1) +45 J
2) -45 J F*d = (0i*3i)+(-15j*3j)+(0k-(-15k)) = (-15*3) = -45
3) +30 J
4) -30 J Formula sheet: Work done by a constant force:
5) +75 J
3) 24 J
A 2.0-kg block slides down a frictionless incline from point A to point B. A force (magnitude P Ke(A)-P * d +mgsin(theta)d
= 3.0 N) acts on the block between A and B, as shown. Points A and B are 2.0 m apart. If the (10)-(3)(2)+(2)(9.8)(sin30)(2) = 23.6J
kinetic energy of the block at A is 10 J, what is the kinetic energy of the block at B?
1) 27 J Not on formula sheet
2) 20 J
3) 24 J
4) 17 J
5) 37 J
The speed of a 4.0-kg object is given by v = (2t) m/s, where t is in s. At what rate is the resultant
e. 16 W
force on this object doing work at t=1 s?
2 steps:
a. 48 W
F=(mass)(v) = 4*2 = F= 8
b. 40 W
W=F*v = 8*2 = 16W
c. 32 W
d 56 W
Not on Formula Sheet:
e. 16 W
1) 3.7 m/s
(1/2)mvi^2 - (1/2)kx^2= v^2
The horizontal surface on which the block slides is frictionless. The speed of the block before Vi=6.0 m/s
it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has X= 15cm = .15m
been compressed 15 cm? k = 2.0 kN/m K=2.00 kNm = 2000 N/m
1) 3.7 m/s m = 2kg
2) 4.4 m/s
3) 4.9 m/s
4) 5.4 m/s 1/2*(2)*(6^2)-(1/2)*(2000)*(0.15^2) = V^2
5) 14 m/s sqrt13.5 = 3.7
Hard to decipher on formula sheet6.0m/s
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, A 0.04-kg ball is thrown from the top of a 30-m tall building (point A) at an unknown angle
a. 12 J
above the top of the buildsing before striking the ground at point B. If air resistance is
negligible, what is the value of the kinetic energy of the ball at B minus the kinetic energy mgh=.04*9.8*30=12J.
of the ball at A (Kb-Ka)?
a. 12 J Formula Sheet: Opposite of Gravitational Potential Energy
b. -12 J
c. 20 J
d. -20 J
e. 32 J
A 0.40-kg particle moves under the influence of a single conservative force. At point A where
the particle has a speed of 10 m/s, the potential energy associated with the conservative force
is +40 J. As the particle moves from A to B, the force does +25 J of work on the particle. What 2) +15 J
is the value of the potential energy at point B?
1) +65 J PEa - Work Done = PE at B When "Conserved and going in the same direction"
2) +15 J
3) +35 J +40-25 = 15 J
4) +45 J
5) -40 J
a. 17m/s
A skier weighing 0.80 kN comes down a frictionless ski run that is circular (R = 30) at the
vf^2 = 2(g)(R)(1-costheta)+Vi^2
bottom, as shown. If her speed is 12 m/s at point A, what is her speed at the bottom of the
Vf^2=2(9.8)(30)(1-cos40)+144
hill (Point B)
SQRT answer = Vf
a. 17 m/s
b. 19 m/s
c. 18 m/s
d. 20 m/s
e. 12 m/s
A 1.2-kg mass is projected from ground level with a velocity of 30 m/s at some unknown angle
above the horizontal. A short time after being projected, the mass barely clears a 16-m tall 1) 0.35
fence. Disregard air resistance and assume the ground is level. What is the kinetic energy of
the mass as it clears the fence?
1) 0.35 kJ 1/2 MV^2 = x + mgh
2) 0.73 kJ
3) 0.40 kJ 540= x+188.35
4) 0.68 kJ K=351.84J Convert to KJ = .35
5) 0.19 kJ
c. -8.1 J
A 1.2-kg mass is projected down a rough circular track (radius = 2.0 m) as shown. The speed
of the mass at point A is 3.2 M/s, and at point B, it is 6. m/s. How much work is done on the 1/2(m) (vf^2-Vi^2)-(mgr)
mass between A and B by the force of friction.
1/2*1.2* (6^2-3.2^2) - (1.2 * 9.8*2)
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2.0 N) acts on the block between A and B, as shown. Points A and B are 1.5 m apart. If the
kinetic energies of the block at A and B are 5.0 J and 4.0 J, respectively, how much work is Wf = (Kef-Kei)-P*costheta*d =
done on the block by the force of friction as the block moves from A to B? (4-5) - 2cos40(1.5) = -3.3 J
1) -3.3 J Not on formula sheet
2) +1.3 J
3) +3.3 J
4) -1.3 J
5) +4.6 J
A constant force of 15 N in the negative y direction acts on a particle as it moves from the origin
to the point (3i+3j-k) m. How much work is done by the given force during this displacement? 2) -45 J
1) +45 J
2) -45 J F*d = (0i*3i)+(-15j*3j)+(0k-(-15k)) = (-15*3) = -45
3) +30 J
4) -30 J Formula sheet: Work done by a constant force:
5) +75 J
3) 24 J
A 2.0-kg block slides down a frictionless incline from point A to point B. A force (magnitude P Ke(A)-P * d +mgsin(theta)d
= 3.0 N) acts on the block between A and B, as shown. Points A and B are 2.0 m apart. If the (10)-(3)(2)+(2)(9.8)(sin30)(2) = 23.6J
kinetic energy of the block at A is 10 J, what is the kinetic energy of the block at B?
1) 27 J Not on formula sheet
2) 20 J
3) 24 J
4) 17 J
5) 37 J
The speed of a 4.0-kg object is given by v = (2t) m/s, where t is in s. At what rate is the resultant
e. 16 W
force on this object doing work at t=1 s?
2 steps:
a. 48 W
F=(mass)(v) = 4*2 = F= 8
b. 40 W
W=F*v = 8*2 = 16W
c. 32 W
d 56 W
Not on Formula Sheet:
e. 16 W
1) 3.7 m/s
(1/2)mvi^2 - (1/2)kx^2= v^2
The horizontal surface on which the block slides is frictionless. The speed of the block before Vi=6.0 m/s
it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has X= 15cm = .15m
been compressed 15 cm? k = 2.0 kN/m K=2.00 kNm = 2000 N/m
1) 3.7 m/s m = 2kg
2) 4.4 m/s
3) 4.9 m/s
4) 5.4 m/s 1/2*(2)*(6^2)-(1/2)*(2000)*(0.15^2) = V^2
5) 14 m/s sqrt13.5 = 3.7
Hard to decipher on formula sheet6.0m/s
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, A 0.04-kg ball is thrown from the top of a 30-m tall building (point A) at an unknown angle
a. 12 J
above the top of the buildsing before striking the ground at point B. If air resistance is
negligible, what is the value of the kinetic energy of the ball at B minus the kinetic energy mgh=.04*9.8*30=12J.
of the ball at A (Kb-Ka)?
a. 12 J Formula Sheet: Opposite of Gravitational Potential Energy
b. -12 J
c. 20 J
d. -20 J
e. 32 J
A 0.40-kg particle moves under the influence of a single conservative force. At point A where
the particle has a speed of 10 m/s, the potential energy associated with the conservative force
is +40 J. As the particle moves from A to B, the force does +25 J of work on the particle. What 2) +15 J
is the value of the potential energy at point B?
1) +65 J PEa - Work Done = PE at B When "Conserved and going in the same direction"
2) +15 J
3) +35 J +40-25 = 15 J
4) +45 J
5) -40 J
a. 17m/s
A skier weighing 0.80 kN comes down a frictionless ski run that is circular (R = 30) at the
vf^2 = 2(g)(R)(1-costheta)+Vi^2
bottom, as shown. If her speed is 12 m/s at point A, what is her speed at the bottom of the
Vf^2=2(9.8)(30)(1-cos40)+144
hill (Point B)
SQRT answer = Vf
a. 17 m/s
b. 19 m/s
c. 18 m/s
d. 20 m/s
e. 12 m/s
A 1.2-kg mass is projected from ground level with a velocity of 30 m/s at some unknown angle
above the horizontal. A short time after being projected, the mass barely clears a 16-m tall 1) 0.35
fence. Disregard air resistance and assume the ground is level. What is the kinetic energy of
the mass as it clears the fence?
1) 0.35 kJ 1/2 MV^2 = x + mgh
2) 0.73 kJ
3) 0.40 kJ 540= x+188.35
4) 0.68 kJ K=351.84J Convert to KJ = .35
5) 0.19 kJ
c. -8.1 J
A 1.2-kg mass is projected down a rough circular track (radius = 2.0 m) as shown. The speed
of the mass at point A is 3.2 M/s, and at point B, it is 6. m/s. How much work is done on the 1/2(m) (vf^2-Vi^2)-(mgr)
mass between A and B by the force of friction.
1/2*1.2* (6^2-3.2^2) - (1.2 * 9.8*2)
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