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A box of mass

2 \mathrm{~kg}

is being pushed horizontally by a $10 \mathrm{~N}...

Jan 29, 2024

A box of mass

2 \mathrm{~kg}

is being pushed horizontally by a

10 \mathrm{~N}

force for

60 \mathrm{~cm}

along the floor. During the first

20 \mathrm{~cm}

, the friction between the box and the floor is

4 \mathrm{~N}

, and the friction for the rest of the distance is negligible. What is the work done on the box by the

10 \mathrm{~N}

force? A

2 \mathrm{~J}

3 \mathrm{~J}

4 \mathrm{~J}

6 \mathrm{~J}

Solution by Steps

step 2

Substitute the given values for the first 20 cm:

W_1 = (10 \text{ N} - 4 \text{ N}) \times 0.2 \text{ m}

step 3

Calculate the work done for the first 20 cm:

W_1 = 6 \text{ N} \times 0.2 \text{ m}

step 4

Calculate the work done for the remaining 40 cm where friction is negligible:

W_2 = F_{\text{applied}} \times d_2

step 5

Substitute the given values for the remaining 40 cm:

W_2 = 10 \text{ N} \times 0.4 \text{ m}

step 6

Calculate the work done for the remaining 40 cm:

W_2 = 10 \text{ N} \times 0.4 \text{ m}

step 7

Add the work done for both segments to find the total work done:

W_{\text{total}} = W_1 + W_2

step 8

Substitute the calculated values for

W_1

and

W_2

W_{\text{total}} = 1.2 \text{ J} + 4 \text{ J}

step 9

Calculate the total work done:

W_{\text{total}} = 5.2 \text{ J}

Key Concept

Work Done by a Constant Force

Explanation

The work done by a constant force is calculated as the product of the force, the distance over which it is applied, and the cosine of the angle between the force and the direction of motion. In this case, the force is horizontal and the motion is horizontal, so the angle is 0 degrees and the cosine of 0 degrees is 1.