IGCSE Physics 0625: Friction, Resultant Force and F = ma for a Dragged Crate

A worker in a workshop sands a wooden plank by pushing a sanding block back and forth across its surface.

(a) State the effect that friction has on the motion of the sanding block as it is pushed across the plank, and state the direction friction acts in, relative to the block's motion. [2]

(b) State one other effect that this friction has, referring to energy. [2]

(c) The worker then drags a wooden crate of mass $25\text{ kg}$ across the workshop floor at a slow, constant velocity, by applying a horizontal pushing force of $85\text{ N}$ . State the size of the friction force acting on the crate, and explain your answer using the resultant force and Newton's first law. [2]

(d) The worker pushes harder, so the pushing force increases to $110\text{ N}$ while the friction force stays at $85\text{ N}$ . Calculate the resultant force now acting on the crate. [1]

(e) Calculate the acceleration of the crate produced by this resultant force. [1]

Part (a): Effect and direction of friction on the sanding block

As the sanding block is pushed across the plank, friction between the two surfaces opposes (impedes) its motion. It makes the block harder to push. This friction force acts in the opposite direction to the block’s motion at every instant, whichever way the block is being pushed.

Part (b): Another effect of friction

Besides opposing motion, friction between the sanding block and the plank transfers energy to heat (thermal energy). This is why both the sanding block and the plank become noticeably warmer as the sanding continues. Some of the work done pushing the block is converted into thermal energy at the rubbing surfaces, rather than useful motion.

Part (c): Friction force while moving at constant velocity

The crate moves at a constant velocity, so by Newton’s first law the resultant force acting on it must be zero. An object continues at a constant velocity only when there is no unbalanced force acting on it.

Since the only horizontal forces on the crate are the $85\text{ N}$ pushing force (forward) and friction (backward), and these must sum to zero:

$\text{friction force} = \text{pushing force} = \boxed{85\text{ N}}$

Part (d): Resultant force after pushing harder

The pushing force and friction now act along the same line but in opposite directions, so subtract to find the resultant:

$\text{Resultant force} = 110\text{ N} - 85\text{ N} = \boxed{25\text{ N}}$

This resultant force acts forward, since the pushing force is now bigger than the friction force.

Part (e): Acceleration of the crate

Use $F = ma$ with the resultant force from part (d) and the crate’s mass of $25\text{ kg}$ :

$F = ma \quad\Rightarrow\quad a = \frac{F}{m}$

$a = \frac{25\text{ N}}{25\text{ kg}}$

$a = \boxed{1.0\text{ m/s}^2}$

Final answers

(a) Friction opposes the block’s motion, acting in the opposite direction to it
(b) Friction transfers energy to heat, warming the block and plank
(c) Friction force $=$ $85\text{ N}$ , since the resultant force is zero at constant velocity (Newton’s first law)
(d) Resultant force $=$ $25\text{ N}$ forward
(e) Acceleration $=$ $1.0\text{ m/s}^2$