1. A 75 kg bobsled is pushed along a horizontal surface by two athletes. After 4
ID: 1991635 • Letter: 1
Question
1. A 75 kg bobsled is pushed along a horizontal surface by two athletes. After 4.5 m from rest, the bobsled has a speed of 6.0 m/s. How much force did the two athletes push with?
2. In 1990, Roger Hickey, 74 kg, of California reached a speed of 35 m/s (77 mph) on his skateboard, 3.6 kg. How much force would be needed to stop him in a distance of 150 m (about half a football field)?
3. A car that weighs 14700 N is traveling at 25 m/s. The brakes are applied and the car stops. The braking force is –7100 N. What was the braking distance?
4. How much work is required to stop an electron, m = 9.31x10-31 kg, which is moving with a speed of 1.90x106 m/s?
5. A weighlifter raises a 180-kg barbell to a height of 1.95 m. What is the increase in the barbell’s potential energy?
6. A rifle can fire a 4.20-gram bullet at a speed of 965 m/s.
(a) How much work is done on the bullet?
(b) If the length of the barrel is 75 cm, how much force accelerates the bullet?
7. At an accident scene on a level road, investigators measure a car’s skid marks to be 88 m long. It is a rainy day. The coefficient of friction between the tires of the car and the road is estimated to be 0.42. Determine the speed of the car just before the driver slammed on the brakes. (Hint: Ff does work to stop the car.)
8. In the 1950s, an experimental train that had a mass of 25000 kg was powered across a level track by a jet engine that produced a thrust of 500000 N for a distance of 500 m. What would be the speed of the train after this distance?
Explanation / Answer
1.work done = avg. force* distance = net change in kinetic energy = force*4.5= 0.5*m*v2
force= 300 newtons
2.force * 150=0.5* 77.6 * 352
force = 91.05 newtons
3. -7100(F)*Dis= 0.5*14700*252/9.81 (mass= weight/g)
dis = 65.95 mts
4. w = 0.5*9.31x10-31 kg*(1.90x106)2
=16.8 * 10-19N
5. change in potential energy= m*g*h= 180-kg*9.81* 1.95 m. = 3443.31J
6. A rifle can fire a 4.20-gram bullet at a speed of 965 m/s.
(a) work done on the bullet = net change in kinetic energy = 0.5*4.2*10-3*9652
= 1955.572J
(b) length of the barrel is 75 cm
avg. force* distance = work done
force = 2607.430N
7. car’s skid marks to be 88 m long. The coefficient of friction 0.42.
Here work is done by friction so frictional force= umg ; u = 0.42
88*9.81*.42= 0.5*v2
v= 26.93 m/s
8. mass of 25000 kg thrust of 500000 N for a distance of 500 m.
thrust(FORCE)*dist= 0.5*m*v2
the speed of the train = 141.42 m/s
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