A paraglider P of mass 95 kg is pulled by a wire attached to a boat, as shown in Fig. 2.1.

Question 28

(a) State the two conditions for a system to be in equilibrium. [2]

(b) A paraglider P of mass 95 kg is pulled by a wire attached to a boat, as shown in Fig. 2.1.

Fig. 2.1

The wire makes an angle of 25° with the horizontal water surface. P moves in a straight line parallel to the surface of the water.

The variation with time t of the velocity v of P is shown in Fig. 2.2.

Fig. 2.2

(i) Show that the acceleration of P is 1.4 m s^-2 at time t = 5.0 s. [2]

(ii) Calculate the total distance moved by P from time t = 0 to t = 7.0 s. [2]

(iii) Calculate the change in kinetic energy of P from time t = 0 to t = 7.0 s. [2]

(iv) The tension in the wire at time t = 5.0 s is 280 N.

Calculate, for the horizontal motion,

1. the vertical lift force F supporting P, [3]

2. the force R due to air resistance acting on P in the horizontal direction. [3]

[Total: 14]

Reference: Past Exam Paper – June 2017 Paper 21 Q2

Solution:

(a)

resultant force (in any direction) is zero

resultant torque/moment (about any point) is zero

(b)

(i)

{We can either use the formula a = v-u / t or the gradient of the graph. The graph at t = 5.0 s is a straight line, so the gradient is constant.}

a = (v − u) / t               or gradient                  or Δv / (Δ)t

{Example: consider the points (7.0, 8.8) and (4.0, 4.6)}

e.g. a = (8.8 − 4.6) / (7.0 – 4.0) = 1.4 m s^-2                                     

(ii)

{The distance travelled is given by the area under the velocity-time graph. The area may be broken down into a rectangle and a trapezium.

Area of rectangle (from t = 0 to t = 4.0 s) = (4.6 × 4) m

Area of trapezium (from t = 4.0 s to t = 7.0 s) = ½ × (8.8 + 4.6) × 3 m

Recall: area of trapezium = ½ × sum of parallel sides × height}

s = (4.6×4) + [(8.8 + 4.6) / 2] × 3

s = 18.4 + 20.1

s = 39 (38.5) m

(iii)

{KE = ½ mv²

We need to consider the speeds at t = 0 s and at t = 7.0 s.}

ΔE = ½ × 95 × [(8.8)² − (4.6)²]

ΔE = 3678 – 1005

ΔE = 2700 (2673) J               

(iv)

1.

{Since the boat is moving horizontally, the resultant vertical force is zero at P.

Forces at P: Weight (down), Vertical component of tension (down), Lift force F (up)

Lift F = Vertical component of tension + Weight

W = mg}

weight = 95 × 9.81 (= 932 N)

vertical tension force = 280 sin 25°    or 280 cos 65° (=118.3 N)

{Lift F = Vertical component of tension + Weight}

F = 932 + 118

F = 1100 (1050) N

2.

{Resultant force = Horizontal tension force – Air resistance R}

horizontal tension force = 280 cos 25° or 280 sin 65° (= 253.8 N)

{Resultant force = ma}

resultant force = 95 × 1.4 (= 133 N)

133 = 253.8 – R

R = 120 (120.8) N