another question:
the 20 kg block is at the top of a 10 m long inclined plane. The block starts from rest and slides without friction down the length of the incline. The block is 5 m high.
1) Determine the gravitational potential energy of the block at the top of the incline.
Will PE = mgh = 20 * 9.8 * 5 work?
2) Determine the kinetic energy of the block when it reaches the bottom of the incline
Will KE = 1/2 mv^2
980 = 1/2 (20) v^2 solve for v work?
physics - bobpursley, Sunday, March 20, 2011 at 8:12pm
I am uncertain of the height on 1). You said the block is 5 m high? Did you mean the inclined plane?
2) again, same issue.
physics - Nabil, Sunday, March 20, 2011 at 9:50pm
Sorry. Yes, the inclined plane is 5 m high.
so will the PE and KE formulas work?
2. There is no friction; therefore, KE = PE = 980 J.
Yes, the formulas for gravitational potential energy (PE = mgh) and kinetic energy (KE = 1/2 mv^2) will work in this scenario. Here's how you can calculate the answers to both questions:
1) Gravitational Potential Energy (PE):
Since the block is at the top of the inclined plane, the height being referred to is the vertical height of the inclined plane, which is 5 m in this case. Therefore, you can use the formula PE = mgh, where m is the mass of the block (20 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (5 m). Substituting the values into the equation:
PE = 20 kg * 9.8 m/s^2 * 5 m = 980 Joules
2) Kinetic Energy (KE):
With no friction acting on the block, all the potential energy at the top of the incline is converted into kinetic energy at the bottom of the incline. Using the formula KE = 1/2 mv^2, where m is the mass of the block (20 kg) and v is the velocity of the block at the bottom of the incline. Rearranging the equation to solve for v, we get:
v = sqrt(2 * KE / m)
To determine the kinetic energy, we need to know the velocity of the block at the bottom of the incline. The information given in the question does not provide the necessary data to calculate the speed.