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Chapter 9: Q. 7 (page 227)

A 20 g particle is moving to the left at 30 m/s. A force on the particle causes it to move the the right at 30 m/s. How much work is done by the force?

Short Answer

Expert verified

Net work done is zero.

Step by step solution

01

Given information

A 20 g particle is moving to the left at 30 m/s.

A force on the particle causes it to move the the right at 30 m/s.

02

Explanation

Work done by force is given by the Change in Kinetic Energy.

Lets assume

vo be the initial velocity = -30 m/s
v1 be the final velocity =+30 m/s

Now find the work done as Change in Kinetic Energy

W=Change in Kinetic Energy

W=12×m×v12-vo2W=12×20×10-3kg×(30m/s)2-(-30m/s)2W=0J

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Most popular questions from this chapter

A 150 g particle at x=0 is moving at 2.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.250N)sin(x/2.00m). What is the particle’s speed when it reaches x=3.14 m?

The two ropes seen in FIGURE are used to lower a 255 kg piano 5.00 m from a second-story window to the ground. How much work is done by each of the three forces?

The rope in FIGURE Q9.4 pulls the box to the left across a rough surface. Is the
work done by tension positive, negative, or zero? Explain.

FIGURE is the force-versus-position graph for a particle moving along the x-axis. Determine the work done on the particle during each of the three intervals 0–1m, 1–2m, and 2–3m.

Hooke’s law describes an ideal spring. Many real springs are better described by the restoring force (Fsp)s=-k∆s-q(∆s)3, where q is a constant.

Consider a spring with k=250N/m .

It is alsoq=800N/m3.

a. How much work must you do to compress this spring 15cm? Note that, by Newton’s third law, the work you do on the spring is the negative of the work done by the spring.

b. By what percent has the cubic term increased the work over what would be needed to compress an ideal spring? Hint: Let the spring lie along the s-axis with the equilibrium position of the end of the spring at s=0.

Then ∆s = s.
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