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Friday, November 23, 2012

Some calcutations on forces and moving objects

I have made a number of mistakes. Your task is to copy and correct my answers as you go.


Laws of Motion


(Use g = 9.8Nkg-1)

 

  1. Find the acceleration of a body of mass 10kg when it is subjected to a horizontal force of 100 N if it (a) can move along a smooth horizontal surface, Ignoring Friction a = F/m = 100/10 = 10ms-2 (b) can move along a horizontal surface which produces a frictional force of 80 N. Resultant Force = 100 – 80 = 20N, using a = F/m = 80/10 = 8 ms-2

 

  1. A rocket of mass 800 000 kg has motors giving a thrust of 9 800000 N. Find the acceleration at lift off Ignoring air resistance a = F/m = 8 x 105 / 9.8 x 106 = 8.1632653061224489795918367346939e-2

 

  1. A force of 100 N acts on a mass of 1 kg. What is the acceleration? Ignoring friction 100ms-1

 

  1. A force of 5 N acts on a stationary mass of 2kg which can move along a smooth horizontal surface. What is its velocity after 5s? Ignoring friction  a = (v-u)/t = F/m. Rearranging v – u = Ft/m As u = 0 v = Ft/m = 5 x 5/2 =12.5 ms-2

 

  1. A mass of 10 kg acquires a velocity of 20 ms-1 from rest in 4 s. What force is required?  Ignoring friction F=m a rearranging F= m (v-u)/t = 10(4-0)/20 = 2N

 

 

  1. A car of mass 600 kg travelling at 72 km h is brought to rest in 54 m after the driver sees an obstruction ahead. If the distance travelled after the driver applies the brakes is 40 m find the driver’s reaction time Distance travelled in time taken for driver to react = 72 – 40 = 32m.  v= x/t so t = x/v = 32/72 = .44 s and the braking force v2 = u2 + 2as  rearranging a= (v2 – u2) / 2s. As F = ma then F = m (v2-u2)/2s = 600(0 – 202)/2 x 40 = 3000N

 

  1. A mass of 2kg projected along a flat surface with a velocity of 15 m s-1 comes to rest after travelling 30 m. What is the frictional force?

 

 v2 = u2 + 2as  rearranging a= (v2 – u2) / 2s. As F = ma then F = m (v2-u2)/2s = 2 x 15/60 = 0.5 N

  1. A Mini of mass 576 kg can accelerate from rest to 72km h-1 in 20 s. If the acceleration is assumed uniform find this acceleration and the tractive force in Newtons needed to produce it.

F=ma = m (v-u)/t = 567 (20 – 0)/20 = 567N

 

  1. A Mini of mass 576 kg can be stopped (in neutral) in 72 m from 108 km h-1 Find (a) the deceleration, v2 = u2 + 2as  rearranging a= (v2 – u2) / 2s. = 1082 / 2 x 72 = 7.5 ms-2(b) the frictional force between the tyres and the road in Newtons, F=ma = 576 x 7.5 = 4328 N

 

  1. The first-stage rocket motors of the Apollo spacecraft produce a thrust of 3.3 x 107 N and the complete spacecraft has a mass of 2.7 x 106 kg. Find (a) the resultant force accelerating the spacecraft,

 

W= mg  = (2.7 x 106) 9.81 = 2.6 x 107. Resultant force = Thrust – weight = (3.3 x 107 - 2.6 x 107 )= 6.5 x 106 N

(b) the initial acceleration, a = F/m = 6.5 x 106 / 3.3 x107 = 0.197 ms-2

(c) the time for it to rise through a distance equal to its own height as it ‘lifts off if its height is 111 m and the average acceleration during this time is 2.5 ms-2
s = ut + ½ at2 As u = 0 then s = ½ at2 rearranging t= √2s/a = √222/2.5 = 5.96 s

  1. A boy of mass 50kg stands in a lift. What will he ‘weigh’ in Newtons if the lift accelerates at 0.50 ms-2 (a) upwards, Force of lift on boy due to acceleration = ma = 50 x 0.5 25N Force of lift on boy due to his weight = 50 x 9.81 = 490.5N total force exerted on lift by boy = 25 + 590.5 = 515.5N  (b) downwards? 465.5N

 
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