A car of mass 1400 kg is travelling on a straight, horizontal road at a constant speed of 25 m s-1. The output power from the car’s engine is 30 kW.

Question 7

A car of mass 1400 kg is travelling on a straight, horizontal road at a constant speed of 25 m s^-1. The output power from the car’s engine is 30 kW.

The car then travels up a slope at 2° to the horizontal, maintaining the same constant speed.

What is the output power of the car’s engine when travelling up the slope?

A 12 kW                      B 31 kW                      C 42 kW                      D 65 kW

Reference: Past Exam Paper – March 2017 Paper 12 Q19

Solution:

Answer: C.

When the car is travelling on a horizontal road, its weight is directly downwards and does not contribute in opposing the motion of the car. 

In terms of energy, the engine is only providing kinetic energy.

Along the slope, a component of the weight opposes the motion of the car. Thus, the car’s engine provides both kinetic energy and gravitational potential energy since the car is gaining in height. In other words, the output power of the car’s engine should now be greater. [A incorrect]

The KE of the car remains the same as the car is moving with constant speed. So, this requires the same amount of power by the car’s engine (= 30 kW).

We need to find out the extra power required by the car on the slope.

Let the distance moved by the car along the slope be h.

Gain in height of the car on the slope is Δh sin 2

Gain in GPE = mgh sin2

Power = Energy / time = (mgh sin 2) / t

Power = mgv sin 2                  since speed v = h / t

Power = 1400 × 9.8 × 25 sin2 = 12.2 kW

This is the extra power required by the car on the slope.

Total power of car’s engine = 30 + 12.2 = 42.2 kW

Atmospheric pressure at sea level has a value of 100 kPa. The density of sea water is 1020 kg m-3. At which depth in the sea would the total pressure be 110 kPa?

Question 5

Atmospheric pressure at sea level has a value of 100 kPa.

The density of sea water is 1020 kg m^-3.

At which depth in the sea would the total pressure be 110 kPa?

A 1.0 m                       B 9.8 m                       C 10 m                        D 11 m

Reference: Past Exam Paper – November 2014 Paper 13 Q20

Solution:

Answer: A.

Let the depth in the sea at which the pressure is 110 kPa be h.

The pressure at h is due to the amount of water above the point and due to the atmospheric pressure at sea level.

Pressure at depth h = Pressure due to height h of sea water + Atmospheric pressure

110 000 = hρg + 100 000

hρg = 10 000

Depth h = 10 000 / (ρg) = 10 000 / (1020×9.81)

Depth h = 0.9993 = 1.0 m

A bicycle brakes so that it undergoes uniform deceleration from a speed of 8 m s-1 to 6 m s-1 over a distance of 7 m.

Question 7

A bicycle brakes so that it undergoes uniform deceleration from a speed of 8 m s^-1 to 6 m s^-1 over a distance of 7 m.

If the deceleration of the bicycle remains constant, what further distance will it travel before coming to rest?

A 7 m              B 9 m              C 16 m                        D 21 m

Reference: Past Exam Paper – November 2012 Paper 13 Q8

Solution:

Answer: B.

Since the deceleration is uniform, we may use the equations of uniformly accelerated motion.

We want to find the further distance travelled before coming to rest.

Initial speed u = 6 m s^-1          

Final speed v = 0 m s^-1

Distance travelled = s = ???

Let the acceleration = a (this will be negative since there is a deceleration)

v² = u² + 2as

We need to find ‘a’ first in order the obtain the further distance travelled ‘s’.

Consider the first part of the motion.

Final speed v = 6 m s^-1 

Initial speed u = 8 m s^-1                        

Distance travelled = 7 m

v² = u² + 2as

Acceleration a = (6² – 8²) / (2×7) = - 2 m s^-2

Now, consider the second part of the motion again.

v² = u² + 2as

Distance travelled s = (0² – 6²) / (2 × -2) = 9 m