How much work must you do to push a 13 kg block of steel across a steel table (muk = 0.6) at a...


How much work must you do to push a 13 kg block of steel across a steel table ({eq}\mu_k {/eq} = 0.6) at a steady speed of 1.0 m/s for 9.7 s?

Work-Energy Theorem

The work-energy theorem in classical physics relates the work done on/by a system to the change in kinetic energy of a system. The work-energy theorem can be stated as:

{eq}\displaystyle W = F_{Net} \: \Delta x = \Delta K {/eq}

Here W is the work done by/on the system, {eq}\displaystyle F_{Net} {/eq} is the net force on the system which acts over distance {eq}\displaystyle \Delta x {/eq}, and {eq}\displaystyle \Delta K {/eq} is the change in kinetic energy of the system.

Answer and Explanation:

Using the work-energy theorem we can say that the total work done is:

{eq}\displaystyle W = F_{Net} \: \Delta x = \Delta K {/eq}

The kinetic energy of the block of steel is {eq}\displaystyle \frac{1}{2} (12 \: \text{kg})(1.0 \: \text{m/s})^2 {/eq} at the start and end of the block's motion since it is moving at a constant speed, so:

{eq}\displaystyle \Delta K = 0 {/eq}


{eq}\displaystyle W = F_{Net} \: \Delta x = 0 {/eq}

So we need to do the same amount of work as the force of friction but in the opposite direction so that the block moves at a constant speed. The force of kinetic friction on the block is:

{eq}\displaystyle F_f = -\mu_k F_N = -\mu_k mg = -(0.6)(13 \: \text{kg})(9.8 \: \text{m}/\text{s}^2) = -7.644 \: \text{N} {/eq}

The force of friction is negative since it opposes the motion and acts over the distance the block covers in 9.7 seconds which is:

{eq}\displaystyle \Delta x = v \Delta t = (1.0 \: \text{m/s})(9.7 \: \text{s}) = 9.7 \: \text{m} {/eq}

Thus the work done by the force of friction is:

{eq}\displaystyle W_f = F_f \Delta x = -74.15 \: \text{J} {/eq}

Thus the work we need to do to oppose this energy loss is:

{eq}\displaystyle W= -W_f = 74.15 \: \text{J} {/eq}

We need to do 74.15 Joules of work to push a 13 kg steel block across a steel table at a constant 1 m/s for 9.7 seconds.

Learn more about this topic:

Practice Applying Work & Kinetic Energy Formulas

from Physics 101: Help and Review

Chapter 17 / Lesson 4

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