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Physics 101
Homework 8
Due February 7
1. A stone is dropped from a height 20 m above the ground. If the initial speed is zero, find
speed of the stone just before it will reach the ground. (Neglect air resistance.)
Solution:
From the conservation of mechanical energy or from the work kinetic energy principle follows:
mv 2
v  20m / s
v  29.8m / s 2 20m   19.8m / s
 mgh  v  2 gh
2
2. A stone is dropped from a height 20 m above the ground with initial velocity 10 m/s directed
down. a) Find speed of the stone just before it will reach the ground. (Neglect air resistance.)
b) What would be your answer if initial velocity is directed up?
Solution:
a) From the conservation of mechanical energy or from the work kinetic energy principle
mv22 mv12
follows:

 mgh  v 2  2 gh  v12
2
2


v2  2 9.8m / s 2 20m  10m / s   22.2m / s
v2  22m / s
2
b) The answer would be the same as above because kinetic energy depends on v 2 . Recall that
that a stone sent up returns to the same point with the same speed (however, with opposite
direction of velocity).
3. A box of mass 5.0 kg is accelerated from rest by a force across frictionless floor at a rate of
2.0 m s 2 for 7.0 s. Find the net work done on the box.
Solution:
Since the acceleration of the box is constant, we can find the distance moved:
2
x  x  x0  v0t  12 at 2  0  12  2.0 m s 2   7 s   49 m .
Then the work done in moving the box is

W  F x cos 0o  max   5 kg  2.0 m s 2
  49 m   4.9 10 J
2
4. (a) If the kinetic energy of an arrow is doubled, by what factor has its speed increased? (b) If
its speed is doubled, by what factor does its kinetic energy increase?
Solution:
(a) Since K  12 mv 2 , then v  2  K  m and so v  K . Thus if the kinetic energy is doubled,
the speed will be multiplied by a factor of
2 .
(b) Since K  12 mv 2 , then K  v 2 . Thus if the speed is doubled, the kinetic energy will be
multiplied by a factor of 4 .
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Physics 101
Homework 8
Due February 7
5. How much work is required to stop an electron m  9.1110 31 kg which is moving with a
speed of 1.90 106 m s ?
Solution:
The work done on the electron is equal to the change in its kinetic energy.


W  K  12 mv22  12 mv12  0  12 9.11 1031 kg 1.90  106 m s

2
 1.64 1018 J
6. The roller-coaster car shown in Fig. is dragged up to point 1 where it is released from rest.
Assuming no friction, calculate the speed at points 2, 3, and 4.
Solution:
Since there are no dissipative forces present, the mechanical energy of the roller coaster will be
conserved. Subscript 1 represents the coaster at point 1, etc. The height of point 2 is the zero
location for gravitational PE. We have v1  0 and y1  35 m .
Point 2:
1
2
mv12  mgy1  12 mv22  mgy2 ; y2  0  mgy1  12 mv22 

v2  2 gy1  2 9.80 m s 2
Point 3:
1
2
  35 m   26 m s
mv12  mgy1  12 mv32  mgy3 ; y3  28 m  mgy1  12 mv32  mgy3 

v3  2 g  y1  y3   2 9.80 m s 2
Point 4:
1
2
  7 m   12 m s
mv12  mgy1  12 mv42  mgy4 ; y4  15 m  mgy1  12 mv42  mgy1 

v4  2 g  y1  y4   2 9.80 m s 2
  20 m   20 m s
2