Question

2. A 7.0 kg box, which is initially at rest at the bottom of a frictionless ramp, is pulled up the ramp by a string

When it reaches a height of 2.0 m, the box is moving at 6.0 m/s. It takes 2.0 seconds for the box to reach this height.
a) Calculate the work done by the string in pulling the box
up
the
ramp.
b) Calculate the power exerted by the string in pulling the box up the ramp.

Answer 1

The work done by the string in pulling the box up the ramp is 263.2 Joules, and the power exerted by the string is 131.6 Watts.

Step-by-step explanation:

Work Done by the String and Power Exerted

To calculate the work done by the string in pulling the box up the ramp, one must consider both the change in potential energy and the kinetic energy the box acquires. The work done by the string (W) can be found by using the formula:

W = ΔKE + ΔPE

Where ΔKE is the change in kinetic energy and ΔPE is the change in potential energy. The kinetic energy (KE) of the box at the top of the ramp is given by:

KE = 0.5 * m * v^2

And the potential energy (PE) at height h is:

PE = m * g * h

By substituting the values provided (m = 7.0 kg, v = 6.0 m/s, h = 2.0 m, and g = 9.8 m/s^2 - the acceleration due to gravity), we find:

ΔKE = 0.5 * 7.0 * 6.0^2

ΔPE = 7.0 * 9.8 * 2.0

The sum of these two gives us the total work done by the string.

Power exerted by the string is defined as the work done divided by the time taken to do that work. Using the formula:

Power = Work / Time

Where time (t) is 2.0 seconds. We can find the power exerted by using the work done from the previous calculation and dividing it by the time.

For part a), let's calculate the work:

• ΔKE = 0.5 * 7.0 * 6.0^2 = 126 Joules
• ΔPE = 7.0 * 9.8 * 2.0 = 137.2 Joules
• W = ΔKE + ΔPE = 126 + 137.2 = 263.2 Joules

For part b), to find the power exerted:

Power = 263.2 Joules / 2.0 seconds = 131.6 Watts

Answer 2

a) The work done in raising the block up to 2.0 m is 137.2 joules

b) The power exerted by the string in pulling the box up the ramp is 68.6 watts

Step-by-step explanation:

The question examines the relationship between work energy and power

The parameters of the pulled box are;

The mass of the box, m = 7.0 kg

The height to which the box is pulled, h = 2.0 m

The velocity at which the box is moving at the 2.0 m height = 6.0 m/s

The time if takes the box to reach the height of 2.0 m = 2.0 seconds

The acceleration due to gravity, 'g' is taken as approximately 9.8 m/s²

a) The work done in raising the block up to 2.0 m = The potential energy gained by the block

The potential energy gained by the block, P.E. = m·g·h

Therefore, we have;

The work done in raising the block up to 2.0 m = P.E. = m·g·h

By substituting the given values, we get;

The work done in raising the block up to 2.0 m = 7.0 kg × 9.8 m/s² × 2.0 m = 137.2 joules

b) Power = The rate of doing work

∴ Power, P = Work done/(Time takes for the work)

The work done in raising the block up to 2.0 m = 137.2 joules

The time if takes the box to reach the height of 2.0 m = 2.0 seconds

∴ The power exerted by the string in pulling the box up the ramp, 'P' is given as follows;

P = 137.2 joules/(2.0 seconds) = 68.6 watts

The power exerted by the string in pulling the box up the ramp, P = 68.6 watts.