Question

A particle of mass 2.0 kg moves under the influence of the force F(x)=(−5x²+7x)N. Suppose a frictional force also acts on the particle. If the particle’s speed when it starts at x=−4.0m is 0.0 m/s and when it arrives at x=4.0m is 9.0 m/s, how much work is done on it by the frictional force between x=−4.0m and x=4.0m?

a) 180 J
b) -180 J
c) 360 J
d) -360 J

Answer 1

The work done on the particle by the frictional force is -180 J.

Step-by-step explanation:

The work done on the particle by the frictional force can be calculated using the work-energy theorem. According to the theorem, the work done on an object is equal to the change in its kinetic energy.

Given that the particle's speed at x = -4.0 m is 0.0 m/s and at x = 4.0 m is 9.0 m/s, we can calculate the change in kinetic energy. The initial kinetic energy is zero, and the final kinetic energy is (1/2)mv², where m is the mass of the particle and v is its final speed.

Substituting the values into the equation, we find that the work done on the particle by the frictional force is -180 J. Therefore, the correct answer is option (b) -180 J.