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Physics 6A
Ch. 7 – In Class Worksheet - Solutions
1) To pass a slow-moving truck, you want your 1300 kg car to speed up from 13.4 m/s to
17.9m/s in 3.0s. How much work is done on the car during this acceleration? What is the
minimum power required for this pass?
2
2

W = ∆K = 1 m v 2f − v i2 = 1 (1300kg) 17.9 m − 13.4 m 
2
2
s
s 

W = 91,553J
(
Power =
)
(
) (
)
Work 91553 J
=
= 30,518 Watts
Time
3 sec
2) An 82.0kg mountain climber is in the final stage of the ascent of 4301-m-high Pikes peak.
What is the change in gravitational potential energy as the climber gains the last 100.0m of
altitude?
We have a formula for potential energy:
∆Ugrav = mg(∆y )
Setting y=0 at the initial height of the climber, we have
yi=0m and yf=100m
Plugging these values in we get:
∆Ugrav = (82kg) 9.8 m2 (100m − 0m) = 80,360J
s 

3) A 1.70kg blocks slides on a horizontal, frictionless surface until it encounters a spring with a
force constant of 955 N/m. The block compresses the spring by 4.60cm before turning around
and sliding in the opposite direction. Find the initial speed of the block. (Ignore air resistance
and any energy lost when the block initially contacts the spring)
Use Conservation of Energy. Initial energy is all
v
initial kinetic, final energy is all elastic potential energy.
E f = Ei
4.6cm
V=0
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final
1 kx 2 = 1 mv 2
2
2
955 N (0.046m)2 = (1.7kg)v 2
m
⇒ v = 1 .1 m
s
(
)
Physics 6A
Ch. 7 – In Class Worksheet (Page 2) - Solutions
4) Deep in the forest, a 17.0g leaf falls from a tree and drops straight to the ground. If its initial
height was 5.30m and its speed on landing was 1.3m/s, how much non-conservative work was
done on the leaf?
We will use conservation of energy. Initially the leaf has gravitational potential
energy, but no kinetic energy (not moving). When it hits the ground it has kinetic
energy. If we define y=0 at the ground, then Ugrav=0 when the leaf hits. So we have
the following values:
Initial: yi=5.3m ; vi=0
Final: yf=0 ; vf=1.3 m/s
5.3m
Our formula is: Ef = Ei + Wother (we need to find W other in this problem)
2
½ mvf = mgyi + W other
2
2
½ (0.017kg)(1.3 m/s) = (0.017kg)(9.8 m/s )(5.3m) + Wother
Wother = -0.869 J (negative work because friction slows the leaf down)
5) When a force of 120.0N is applied to a certain spring, it causes a stretch of 2.25cm. What is
the potential energy of this spring when it compressed by 3.50cm?
First we need to find the spring constant k. We use the force
formula Fspring=kx
120N = (k)(0.0225m) → k = 5333 N/m
Now we can find the potential energy:
Uelastic = ½ kx2
Uelastic = ½(5333 N/m)(0.035m)2
Uelastic = 3.27 J
6) A 100kg crate is pulled along a rough level floor at constant speed by a rope inclined at an
angle of 30°. If the coefficient of kinetic friction is 0.4, find the tension in the rope and the work
done by the rope’s tension when the crate moves 10m.
This problem is mainly a review problem for chapter 5, with a bit of chapter 7 thrown in at the end.
We can start by drawing the free-body diagram for the crate, and setting up the 2 versions of Newton’s 2
Normal
friction
Tension
mg
∑ Fx = 0
Tx − fkinetic = 0
Law.
∑ Fy = 0
Constant speed
means Fnet=0
N + Ty − mg = 0
T ⋅ cos(30 o ) = µ k ⋅ N
N = mg − T sin(30 o )
(
T ⋅ cos(30 o ) = µ k ⋅ mg − T sin(30 o )
(
nd
)
)
Substitute for N in
the x-equation.
T cos(30 o ) + µ k ⋅ sin(30 o ) = µ k ⋅ mg
T=
µ k ⋅ mg
cos(30 o ) + µ k ⋅ sin(30 o )
= 368N
Work = F ⋅ d ⋅ cos θ ⇒ W = (368N) ⋅ (10m) ⋅ cos(30 o ) = 3185J
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