Force Work Energy problems Answers PDF

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EDS: Introductory Physics

FORCES WORK & ENERGY PROBLEMS NUMERICAL ANSWERS ONLY Q5

What resultant force is required to accelerate a 3 kg object by 6 m s-2? F = ma = 3 x 6 =18 N

Q6

A resultant force of 500 N is used to accelerate an object of mass 20 kg. What is the acceleration of the object? F = ma therefore a = F/m = 500/20 = 25 m s-2

Q7

A car of mass 500 kg is travelling along a flat road. The forward force provided between the car tyres and the road is 300 N and the air resistance is 200 N. Calculate the acceleration of the car. The resultant force is 300 – 200 = 100 N (the air resistance opposes the forward motion). a = F/m = 100/500 = 0.2 m s-2

Q8

The maximum forward force a car can provide is 500 N. Air resistance F which the car experiences depends on its speed according to the equation F = 0.2 v2 where v is the speed in m s-1. Determine the top speed of the car. The maximum speed is when the forward force of the car is equal to the air resistance. Therefore 500 = 0.2 v2 rearranging v = √500/0.2 = 50 m s-1

Q9

An aircraft of total mass 1.5 x 105 kg accelerates at maximum thrust from rest along a runway for 25 s before reaching the required speed for take-off of 65 m s-1. Calculate: (a) the acceleration of the aircraft (b) the force acting on the aircraft to produce this acceleration (c) the distance travelled by the aircraft before take-off. (a) We have u = 0, v = 65 m s-1, t = 25 s and we need ‘a’ Use v = u + at therefore 65 = 0 + 25a, a = 65/25 = 2.6 m s-2 (b) F = ma = 1.5 x 105 x 2.6 = 3.9 x 105 N (c) Use s = ½ (u + v)t = ½ (0 + 65) x 25 = 812.5 m We could have also used s = ut + ½ at2 i.e. s = ½ x 2.6 x 252 = 812.5 m

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EDS: Introductory Physics

Q10 What has more kinetic energy, a car of mass 500 kg travelling at 15 m s-1 or a motorcycle of mass 250 kg travelling at 30 m s-1? Kinetic energy given by KE = ½ mv2 Car

KE = ½ x 500 x 152 = 56,250J

Motorcycle

KE = ½ x 250 x 302 = 112,500J

The motorcycle for the reason that the square of the velocity dominated the calculation. Q11 Calculate the change in kinetic energy of a ball of mass 200 g when it bounces. Assume it hits the ground with a speed of 15.8 ms -1 and leaves it at 12.2 m s-1, also where has the energy gone? Mass of ball = 200g = 200/1000 = 0.2kg Down

KE = ½ x 0.2 x 15.82 = 24.964 J

Up

KE = ½ x 0.2 x 12.22 = 14.884 J

Energy lost = 24.964 - 14.884 = 10.08J. The KE is lost as heat to the ground i.e. the temperature of the ground would increase. Q12 A resultant force of 50 N moves an object 3 m, what is the work done? Work done = Fs = 50 x 3 = 150 J Q13 A lift motor provides a force of 20 kN; this force is enough to raise the lift by 18 m in 10 s. Calculate: (a) the work done by the motor (b) the output power of the motor (c) what, if anything, can you say about the electrical input power to the motor? (a) Work done = Fs = 20,000 x 18 = 360,000 J (Note the conversion of kN to N) (b) Output power = work done/time = 360000/10 = 36,000 W i.e. 36kW (c) Because no transformation of energy is 100% efficient, the input power would be greater than 36kW. Q14 A ball of mass 0.5 kg moving at 8 m s-1 rolls over a patch of rough ground and slows to 4 m s-1. What is the work done in slowing the ball? Work done cause the change in kinetic energy Fs = ½ mv2 - ½ mu2 = ½ x 0.5 x 82 - ½ x 0.5 x 42 = 16 – 4 = 12 J

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