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Question: A 10 kg mass moves along x-axis. Its acceleration as function of its position is shown in the figure...

A 10 kg mass moves along x-axis. Its acceleration as function of its position is shown in the figure. What is the total work done on the mass by the force as the mass moves from x = 0 to x = 8 cm?

A

8×10−2J8 \times 10 ^ { - 2 } J

B

C

D

Answer

8×10−2J8 \times 10 ^ { - 2 } J

Explanation

Solution

The total work done on the mass by the force as it moves from x=0x = 0 to x=8x = 8 cm can be calculated by finding the area under the acceleration-position graph and then multiplying by the mass. The area under the graph is a triangle with base 8 cm and height 20 cm/s². Converting to SI units and calculating the work done:

W=12⋅base⋅height⋅m=12⋅(0.08 m)⋅(0.20 m/s2)⋅(10 kg)=0.08 J=8×10−2 JW = \frac{1}{2} \cdot \text{base} \cdot \text{height} \cdot m = \frac{1}{2} \cdot (0.08 \text{ m}) \cdot (0.20 \text{ m/s}^2) \cdot (10 \text{ kg}) = 0.08 \text{ J} = 8 \times 10^{-2} \text{ J}