A trolley moves down a slope, as shown in Fig.1. Slope makes an angle of 25° with the horizontal.

Q#692: [Kinematics + Dynamics] (Past Exam Paper – November 2014 Paper 21 Q4)

A trolley moves down a slope, as shown in Fig.1.

Slope makes an angle of 25° with the horizontal. A constant resistive force F_R acts up the slope on trolley.

At time t = 0, trolley has velocity v = 0.50 m s⁻¹ down the slope.

At time t = 4.0 s, v = 12 m s⁻¹ down the slope.

(a)

(i) Show that acceleration of the trolley down the slope is approximately 3 m s⁻².

(ii) Calculate distance x moved by the trolley down the slope from time t = 0 to t = 4.0 s.

(iii) On Fig.2, sketch variation with time t of distance x moved by the trolley.

(b) Mass of the trolley is 2.0 kg.

(i) Show that component of the weight of the trolley down the slope is 8.3 N.

(ii) Calculate resistive force F_R.

Solution 692:

(a)

(i)

Acceleration a = (v – u) / t or (12 – 0.5) / 4

Acceleration a = (12 – 0.5) / 4 = 2.9 (2.875) (= approximately 3 m s^–2)

(ii) {Distance travelled = Average speed × time}

Distance x = (u + v) t / 2 = [(12 + 0.5) × 4] / 2 = 25 m

(iii) A line with increasing gradient is drawn with a non-zero gradient at origin.

{The gradient of a distance-time graph gives the velocity.

There is a constant acceleration on the trolley, so its speed increases with time. That is, the distance travelled in a unit time becomes greater with time. This is represented by an increase in gradient of the graph.

Now, initially, the speed of the trolley is 0.5ms^-1. So, the gradient cannot be zero at the origin.}

(graph can be drawn better)

(b)

(i) Weight down slope = 2 × 9.81 × sin 25° = 8.29 / 8.3 N

(ii)

(Resultant force F = ma)

8.3 – F_R = 2 × 2.9

Resistive force F_R = 2.5 N (2.3N if 3 used for a)

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