Forces, Movement, Shape, and Momentum
6 marks · NormalTwo students each pull on a rope in opposite directions. Student A pulls with 160 N to the right, Student B pulls with 120 N to the left. The rope is attached to a box of mass 10 kg resting on a rough surface.
Calculate the resultant (unbalanced) force on the box.
Determine the acceleration of the box. (Assume friction is 10 N, acting to the left.)
Draw a vector diagram to show all the horizontal forces acting on the box.
State whether the box moves left, right or remains at rest. Explain why.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) Resultant force = 160 N (right) - [120 N (left) + 10 N (friction, left)] = 30 N (right) N
(b) Acceleration = 3.0 m/s² (to the right) m/s²
(c) Vector diagram shows: 160 N arrow (right), 120 N and 10 N arrows (left), resultant 30 N arrow (right) · must show Individual force arrows (sizes correct and labelled; direction correct; resultant marked)
(d) The box moves to the right because the total force on it is to the right.
Common mistakes: Omitting friction; Incorrect direction for resultant
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Results net force correctly with all forces and directions accounted for.
M2 (b) · 1 mark · Uses F = ma and substitutes values from (a). (follow-through)
A1 (b) · 1 mark · Correct answer for acceleration with correct unit and sign. (follow-through)
D1 (c) · 2 marks · Labels and draws all vectors to scale and in correct direction.
A2 (d) · 1 mark · Correct conclusion with a reason (direction of net force).
Worked solution
a) Right force: 160 N, Left: 120 N (student) + 10 N (friction) = 130 N Resultant = 160 N – 130 N = 30 N (right)
b) a = F / m = 30 N / 10 kg = 3.0 m/s² (right)
d) The box moves to the right because the net force is to the right.