A man travels on a toboggan down a slope covered with snow from point A to point B and then to point C. The path is illustrated in Fig. 1.1.

Question 9

(a) Define acceleration. [1]

(b) A man travels on a toboggan down a slope covered with snow from point A to point B and then to point C. The path is illustrated in Fig. 1.1.

Fig. 1.1 (not to scale)

The slope AB makes an angle of 40° with the horizontal and the slope BC makes an angle of 20° with the horizontal. Friction is not negligible.

The man and toboggan have a combined mass of 95 kg.

The man starts from rest at A and has constant acceleration between A and B. The man

takes 19 s to reach B. His speed is 36 m s^-1 at B.

(i) Calculate the acceleration from A to B. [2]

(ii) Show that the distance moved from A to B is 340 m. [1]

(iii) For the man and toboggan moving from A to B, calculate

1. the change in kinetic energy, [2]

2. the change in potential energy. [2]

(iv) Use your answers in (iii) to determine the average frictional force that acts on the

toboggan between A and B. [2]

(v) A parachute opens on the toboggan as it passes point B. There is a constant deceleration of 3.0 m s^-2 from B to C.

Calculate the frictional force that produces this deceleration between B and C. [2]

 [Total: 12]

Reference: Past Exam Paper – June 2016 Paper 22 Q1

Solution:

(a)

Acceleration is defined as the rate of change of velocity.

(b)

(i)

{v = u + at       where u = 0 (at rest)}

v = 0 + at         or v = at                      

a = 36 / 19 = 1.9 (1.8947) m s^-2            

(ii)

{s = average speed × time      v² = u² + 2as                           s = ut + ½ at² }

s = ½(u + v)t                           or s = v² / 2a                           or s = ½ at²

   = ½×36×19                              = 36² / (2×1.89)                     = ½×1.89×19²

   = 340 m (342 m / 343 m / 341 m)

(iii)

1.

{ΔKE = ½ m (v² – u²)}

ΔKE   = ½×95×(36² – 0²) 

            = 62 000 (61 560) J   

2.

{ΔPE   = mgΔh                       where Δh is the vertical height = 340 sin 40°}

ΔPE   = 95×9.81×340 sin 40°        or 95×9.81×218.5

= 200 000 J                                                                

(iv)

{As the toboggan moves down the slope,

PE at A = KE at B + Work done against friction}

work done (by frictional force) = ΔPE – ΔKE

or

work done = 200 000 – 62 000           (values from 1b(iii) 1. and 2.)            

{Work done = 138 000 N

Work done = (Frictional) Force × distance}

Frictional force = (138 000 / 340 =) 410 (406) N

(v)

{Since there is a deceleration, the resultant force is upwards.

Resultant force = Frictional force – Component of Weight along slope

OR – Resultant force = Component of Weight along slope – Frictional force

Component of Weight along slope = mg sin 20°}

–ma = mg sin 20° – f                          or ma = –mg sin 20° + f

–95 × 3.0 = 95 × 3.36 – f

Frictional force f = 600 (604) N