What is the speed of a 20kg object right before it hits the ground if it was dropped from a height of 30 meters? (Hint: use law of conservation of energy, What is potential energy at the top equal to?)
Answers
v^2-u^2 = 2as
v^2 = 2× 10 × 30
v = √ 600
v = 10√6m\ s.
Answer:
a) Again we define the zero of potential energy to be the lowest point in the problem, where the cannonball is initially fired. Then we can use the height of the cannonball above the ground to determine the potential energy from the formula:
PE = mgh = (5 kg)(10 m/s2)(50 m) = 2500 J
As in the problem above, when the cannonball hits the ground, all of the potential energy it had at the maximum height is transformed into kinetic energy. Using the same idea of energy conservation, we relate the potential energy lost by the cannonball to the kinetic gained: PE top = KE bot.
PE lost = KE gained= 1/2 mv2
½ (5 kg) v2 = 2500 J
v2= 1000 m2/s2 v= 31.62 m/s
EX. #3: A 2.5 kg bowling ball rolls into a wall traveling at 6 m/s. It crashes into the wall and is brought to rest by the forces acting from the wall.
a) How much work was done by the wall in bringing the object to rest?
b) If the ball comes to rest in 0.4 seconds, then calculate the power used in stopping the ball.
SOLUTION:
a) Since the bowling ball stops after hitting the wall, all of the kinetic energy it had initially is removed by the work due to forces by the wall. In this case the work done is equal to the change in kinetic energy of the ball, W=KE:
W= KE = 1/2mv2
W = ½(2.5 kg)(6 m/s)2 = 45 J
b) The power is defined as the rate of change in work or energy P = (Work Energy)/time. The energy used to stop the ball was found above from the change in kinetic energy, or work done by the wall in bringing the bowling ball to rest. Using the time to stop the ball in addition to the power formula, the power can be calculated as
P = (Work Energy)/time = 45 J/0.4 s = 112.5 W