Can you answer the questions and show your work to the problem

Question

A small block, with a mass of 4 kg, starts from rest at the top of the apparatus shown above. It then slides without friction down the incline, across the first level section, around the loop and then onto the final level section on the right. The maximum height of the incline is 5 m, and the radius of the loop is 1.0 m.

1.Find the initial potential energy of the block at point A.

2.Find the velocity the block at the bottom of the incline at point B.

3.Find the velocity of the block at the top of the loop at point C.

Answer 1

Hi Gavin! This problem deals primarily with the conservation of energy. Because there is no friction, we can write our energy equation as the following:

PE_i + KE_i = PE_f + KE_f

Here, potential energy is equivalent to mgh, so we can use that to find the initial potential energy of the block at point A:

mass of block = 4 kg, h = 5 m, and g = 9.8 m/s²

Therefore, PE of the block at point A = 4*5*9.8 = 196 J.

Next, we can use our conservation of energy equation to determine the velocity of the block at the bottom of the incline.

PE_i + KE_i = PE_f + KE_f

Since the block starts from rest, KE_i = 0. Similarly, once the block is at the bottom of the incline, its final potential energy is equivalent to 0, because h = 0. Therefore, our simplified conservation equation would look like the following:

196 J + 0 = 0 + KE_f

Our final kinetic energy is 196 J. Using the equation KE = mv²/2, we can solve for the speed by substituting 4 kg for the mass:

196 = 4 * v² / 2

v² = 98

v = 9.90 m/s

The velocity of the block at the bottom of the incline is therefore 9.90 m/s.

Finally, we can use the same conservation of energy equation to solve for the velocity of the block at the top of the loop. Because the loop has r = 1.0 m, that means that at position C, the top of the loop, the block is 2r above the ground, or 2 * 1.0 m = 2.0 m.

We can choose any position to be our starting point, so for simplicity's sake, I will use the original starting position of the block at the top of the incline.

PE_i + KE_i = PE_f + KE_f

After substituting known values, we arrive at the following equation:

196 J + 0 = mgh_f + mv_f²/2

Since we know m, g and h_f = 2.0 m, we can substitute these in as well:

196 J = 4*9.8*2 + 4*v_f²/2

2*v_f² = 196 J - 96.04 J = 99.96 J

v_f² = 49.98 m²/s²

v_f = 7.07 m/s

For the block at the top of the loop (point C), the velocity has a magnitude of 7.07 m/s.

To recap answers:

1). 196 J

2). 9.90 m/s

3). 7.07 m/s