Resolving Forces (answers)

Resolving Forces

1. A weight of 50.0 N is suspended from a beam by a string. Calculate the horizontal force must be applied to the weight to keep the string at an angle of 40° to the vertical. 65.3N  Calculate the tension in the string. 
2. A weight of 20 N rests on a plane inclined at 40° to the horizontal. Calculate the components of the weight parallel to the plane and perpendicular to the plane.12.86N, 15.32N
3. A weight of 4.33 N is suspended by a string fastened at its upper end. A horizontal force is applied to the weight so that the string makes an angle of 30⁰ with the vertical. Calculate the force and the tension in the string.2.50N, 5.00N
4. A 10 N weight rests on a smooth inclined plane. A force of 5N parallel to the plane is needed to prevent the weight slipping down the plane. Calculate the reaction of the plane and the angle the plane makes with the horizontal.8.66N ,30⁰
5. A garden roller of weight 800 N is being pulled along by a force of 200 N at 40° to the ground. Calculate (a) the force pulling the roller forwards, 153.2N(b) the vertical force exerted by the roller on the ground. 671.4 N
6. Calculate the forces (a) 153.2N and (b) 928.6N in the previous problem if the roller is being pushed instead of being pulled.

Mark Scheme for Cirular motion 3 Conical Pendulum

1 Explanation

There is a resultant (or net or unbalanced) force

􀀹

Plus any 3 of following:-

Direction of motion is changing 􀀹

Velocity is changing 􀀹

Velocity change implies acceleration 􀀹

Force produces acceleration by F = ma (or N2) 􀀹

Force (or acceleration) is towards centre / there is a centripetal

force (or acceleration) / no force (or acceleration) parallel to

motion

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No work done, so speed is constant 􀀹 Max

Newton's secon law answers

Newton’s Second Law of Motion
(Use g = 9.8 N kg^-1)

Worked example

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, (b) can move along a horizontal surface which produces a frictional force of 80N.
(a) F=ma     

Rearranging

 a = F/m                F force in N            m mass in kg          a acceleration in ms-2
 a = 100 / 10  = 10 ms^-2

(b) The resultant force = 100 N – 80 N = 20 N

 

F=ma

rearranging a = F/m
 a = 20 /10
a = = 2ms^-2  

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

a = F/m = 100/ 1 = 100 ms^-2

2. A mass of 10kg acquires a velocity of 20 ms^-1 from rest in 4 s. a = (v-u) / t = (20 – 0) /4 = 5 ms^-2 Calculate the force is required. F = ma = 10 x 5 = 50N
3. A rocket of mass 800 000 kg has motors giving a thrust of 9 800 000 N. Upward force = thrust – weight = 9.8 x 10⁶  - (8 x 10⁵)9.8 =  1.96 x10⁶ N

Calculate the acceleration at lift off. a = F/m = 1.96 x10⁶ / 8 x10⁵ = 2.45 ms^-2

4. A force of 5 N acts on a stationary mass of 2kg which can move along a smooth horizontal surface.

a = F/m = 5/2 = 2.5 ms^-2

Calculate its velocity after 5 s?

v = u +at v = 0 + 2.5 x 5 = 12.5 ms^-1

5. A car of mass 600 kg travelling at 72km h^-1 ( 72000m / 36000 s = 20 ms^-1) 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 40m calculate the driver’s reaction time using v = x/t rearranging t =x/v = (54 - 40) / 20 = 0.7s   and the braking force.  Using v² = u² + 2as (v = 0, u = 20ms^-1, s = 40m ) rearranging 2as = v² – u² and

a = (v² – u²)/ 2s, a = (0 – 20²)/ 80 = 400/80 = 5 ms^-2.

F = ma = (600kg)5ms^-2 = 3000N

6. A mass of 2kg projected along a flat surface with a velocity of 15 ms^-1 comes to rest after travelling 30 m. Calculate the frictional force? Using v² = u² +2as (v = 0, u = 15ms^-1, s = 30m)

rearranging a = (v² – u²)/ 2s, a = (0 -152) / 60 = 3.75 ms^-2

F=ma = 2 x 3.75 = 7.5 N

7. A Mini of mass 576 kg can accelerate from rest to 72 km h^-1 in 20 s. ( 72000m / 36000 s = 20 ms^-1) If the acceleration is assumed uniform calculate this acceleration and the tractive force in Newtons needed to produce it. Using a = (v-u) /t , a = (20-0)/20 = 1ms^-2.

F=ma = 576 x 1 = 576N

A Mini of mass 576 kg can be stopped (in neutral) in 72 m from 108 kmh^-1.(108000/3600 = 30ms^-1) Calculate (a) the deceleration, Using v² = u² + 2as (v = 0, u = 30ms^-1, s = 72m )

rearranging 2as = v² – u² and a = (v² – u²)/ 2s, a = (0 – 30²)/ (2 x 72) = 6.25 ms^-2

(b) the frictional force between the tyres and the road in Newtons. F=ma = 576 x 6.25 = 3600N

8. 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. Calculate (a) the resultant force accelerating the spacecraft, resultant force = thrust – weight = 3.3 x 10⁷ N – (2.7 x 10⁶ x 9.8) = 6.54 x10⁶N

(b) the initial acceleration, F=ma rearranging a = F/m = 6.54 x 10⁶ N / 2.7 x10⁶ kg = 2.42ms^-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

Using s = ut + ½ at² , s = 111m, u = 0 a = 2.5 s = 0 + ½ at² , t² = 2s/a, t= √2s/a = √222/2.5 = 9.42s

9. A boy of mass 50kg stands in a lift. What will he ‘weigh’ in Newtons if the lift accelerates at 0.50 ms^-2

‘Weight’ is caused by the normal reaction force which acts upwards

(a)  upwards, lift supplies normal reaction due to weight of boy + force to accelerate boy upwards

net force = m(g + a) = m (9.8 +0.5)= 515N

(b) downwards? lift supplies normal reaction due to weight of boy - force to accelerate boy downwards (which is ‘supplied’ by gravity)

net force = m(g -a) = m (9.8 -0.5)= 465N