Parachute

Questions on Electric Fields (in no particular order)

E Field

Parachute

Free body Force

Force in E Field

Ping Pong Balls

Projectile Motion

Force between point charges

1. 17983 N - very high charges
2. 3.68 x 10^-9 N
3. 1.63 N
4. 28.4 deg

Newton's second law of Motion

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

Experiment scales and Charged Bodies

Graitaional Potential

.         (a)     force (of attraction) on unit mass (at that point in space/at the
surface of a planet) (1)                                                                                        1

  (b)   (i)      (mgh =) 1500 × 40 × 1.5 × 105; = 9.0 × 109 (J)                                           2

(ii)     (larger as) g decreases with height                                                            1

(iii)     v = 2πR/T; = 2π × 2.0 × 107/(4.5 × 103) = 2.8 × 104 m s–1 (2)
½ mv2; = 0.5 × 1500 × (2.8 × 104)2 = 5.9 × 1011 (J) (2)                             4
aliter: F = mv2/R; = mg; so ½ mv2 = ½ mgR;= 6.0 × 1011 (J)

Sheet on Projectiles 6

3 questions on 1 sheet

satellite and earth