# CIE A level Math -Mechanics : 4.5 Energy, work and power: work done by a force: Exam Style Questions Paper 4

### Question

A block B of mass 2.7 kg is pulled at constant speed along a straight line on a rough horizontal floor. The pulling force has magnitude 25 N and acts at an angle of θ above the horizontal. The normal component of the contact force acting on B has magnitude 20 N.

(i) Show that sin θ = 0.28. [2]

(ii) Find the work done by the pulling force in moving the block a distance of 5 m. [2]

(i) [20 + 25sinθ = 2.7g]

sinθ = 0.28

(ii) [25 × 5 × √(1 – 0.282)]

Work done is 120 J

### Question

A block B lies on a rough horizontal plane. Horizontal forces of magnitudes 30 N and 40 N, making angles of α and β respectively with the x-direction, act on B as shown in the diagram, and B is moving in the x-direction with constant speed. It is given that cos α = 0.6 and cos β = 0.8.

(i) Find the total work done by the forces shown in the diagram when B has moved a distance of 20 m. [2]

(ii) Given that the coefficient of friction between the block and the plane is $$\frac{5}{8}$$ , find the weight of the block. [3]

(i) [WD = 30 × 20 × 0.6 + 40 × 20 × 0.8]

Work done is 1000 J

(ii) 30 × 0.6 + 40 × 0.8 – 0.625W = 0

Weight is 80 N

### Question

The diagram shows the vertical cross-section ABCD of a surface. BC is a circular arc, and AB and CD are tangents to BC at B and C respectively. A and D are at the same horizontal level, and B and C are at heights 2.7 m and 3.0 m respectively above the level of A and D. A particle P of mass 0.2 kg is given a velocity of 8 m$$s^{−1}$$ at A, in the direction of AB (see diagram). The parts of the surface containing AB and BC are smooth.
(i) Find the decrease in the speed of P as P moves along the surface from B to C.
The part of the surface containing CD exerts a constant frictional force on P, as it moves from C to
D, and P comes to rest as it reaches D.
(ii) Find the speed of P when it is at the mid-point of CD.

(i) 1⁄2 mv$$_B^{2}$$
= 1⁄2$$mv_A^{2}$$– mg × 2.7and 1⁄2 m$$v_c^{2}$$
= 1⁄2 mv$$_A^{2}$$
– mg × 3

$$[v_{B}^{2}$$=\ (8^{2}-20\times 2.7,v_{c}^{2}\)= $$8^{2}-20\times 3]$$
loss of speed=$$10^{\frac{1}{2}-2$$=$$1.16ms^{-1}}$$

(ii) Work done = 1⁄2 0.2 × 2 2 + 0.2 × g × 3

1⁄2 (0.4 + 6) = $$1/20.2v_{M}^{2}+0.2g\times 1.5$$

Speed at midpoint is 1.41 m$$s^{–1}$$