Question

A block of mass m
slides up and down over a series of smooth icy hills. The top of the first hill is 7.0 m high from the ground level and the speed of the block at the top of the hill is 15m/s.
Determine speed of the block when it is at the top of the following hill which is 3.0 m high from the zero level.

A) 17 m/s

B) 14m/s

C) 10 m/s

D) 35 m/s

Answer 1

Answer 2

To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy of the block, consisting of kinetic energy (KE) and potential energy (PE), remains constant as it slides up and down the hills, assuming no non-conservative forces (such as friction) are acting.

The total mechanical energy (E) of the block at any point can be expressed as:

E = KE + PE

Given that the block slides up and down the hills smoothly, there is no loss of mechanical energy. Therefore, the total mechanical energy at the top of the first hill is equal to the total mechanical energy at the top of the second hill.

1. At the top of the first hill (height = 7.0 m):

Total mechanical energy at the first hill top (E1) = KE1 + PE1 KE1 = (1/2) * m * v1^2 (where v1 is the speed of the block at the top of the first hill) PE1 = m * g * h1 (where h1 is the height of the first hill)

2. At the top of the second hill (height = 3.0 m):

Total mechanical energy at the second hill top (E2) = KE2 + PE2 KE2 = (1/2) * m * v2^2 (where v2 is the speed of the block at the top of the second hill) PE2 = m * g * h2 (where h2 is the height of the second hill)

Since the mechanical energy is conserved, E1 = E2:

(1/2) * m * v1^2 + m * g * h1 = (1/2) * m * v2^2 + m * g * h2

Now, plug in the given values: v1 = 15 m/s (speed at the top of the first hill) h1 = 7.0 m (height of the first hill) h2 = 3.0 m (height of the second hill) g ≈ 9.81 m/s^2 (acceleration due to gravity)

(1/2) * m * (15 m/s)^2 + m * 9.81 m/s^2 * 7.0 m = (1/2) * m * v2^2 + m * 9.81 m/s^2 * 3.0 m

Now, solve for v2:

112.5 m^2/s^2 + 68.67 m^2/s^2 = (1/2) * v2^2 + 29.43 m^2/s^2

181.17 m^2/s^2 = (1/2) * v2^2

v2^2 = 2 * 181.17 m^2/s^2

v2 ≈ √(2 * 181.17) m/s ≈ 17 m/s

So, the speed of the block when it is at the top of the second hill (3.0 m high) is approximately 17 m/s. Therefore, the correct answer is:

A) 17 m/s