Question

A man weighing 180 lbf pushes a block weighing 100 lbf along a horizontal plane. The dynamic coefficient of friction between the block and plane is \(0.2 .\) Assuming that the block is moving at constant speed, calculate the work required to move the block a distance of \(100 \mathrm{ft}\) considering \((a)\) the man and \((b)\) the block as the system. Express your answers in both lbf-ft and Btu.

Short answer

Answer: The work done when considering the man and the block as the system is 2000 lbf-ft or 2.57 Btu.

Step-by-step solution

Calculate the friction force

The friction force (F_friction) can be calculated using the formula: \(F_{friction} = µ * F_{normal}\) where \(µ\) is the dynamic coefficient of friction and \(F_{normal}\) is the normal force acting on the block which is equal to the weight of the block. \(F_{normal} = 100 \mathrm{lbf}\) (since the weight of the block is provided) Now, we can calculate the frictional force: \(F_{friction} = 0.2 * 100 = 20 \mathrm{lbf}\)

Calculate the force exerted by the man

As the block is moving at a constant speed, the net force acting on it is zero. Thus, the force exerted by the man (F_man) should be equal to the friction force: \(F_{man} = F_{friction} = 20 \mathrm{lbf}\)

Compute the work done for (a) the man and (b) the block as the system

The work done (W) is the product of the force exerted and the distance traveled. In this case: \(W = F_{man} * d\) where d is the distance traveled (100 ft). (a) When considering the man as the system: \(W_{man} = F_{man} * d = 20 \mathrm{lbf} * 100 \mathrm{ft} = 2000 \mathrm{lbf-ft}\) To convert this to Btu, we use the conversion factor: \(1 \mathrm{Btu} = 778.169 \mathrm{lbf-ft}\) \(W_{man} = \frac{2000}{778.169} \approx 2.57 \mathrm{Btu}\) (b) When considering the block as the system: \(W_{block} = F_{man} * d = 20 \mathrm{lbf} * 100 \mathrm{ft} = 2000 \mathrm{lbf-ft}\) (the work done remains the same) Converting this to Btu: \(W_{block} = \frac{2000}{778.169} \approx 2.57 \mathrm{Btu}\) Thus, the work done when considering (a) the man and (b) the block as the system is \(2000 \mathrm{lbf-ft}\) or \(2.57 \mathrm{Btu}\).