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M1 forces question

Hi could i have some help with this question please.

I think a is 0.7ms^-2 and b is 560N

A car which has run out of petrol is being towed by a breakdown truck along a straight horizontal road. The truck has a mass 1200kg and the car has mass 800kg. The truck is connected to the car by a horizontal rope which is modelled as light and inextensible. The truck's engine provides a constant driving force of 2400N. The resistances to motion of the truck and the car are modelled as constant and of magnitude 600N and 400N respectively. Find

a) the acceleration of the truck and the car

b) the tension in the rope

When the truck and the car are moving at 20ms^-1, the rope breaks. The engine of the truck provides the same driving force as before. The magnitude of the resistances to the motion of the truck remains 600N

c) Show that the truck reashes a speed of 28ms^-1 approx 6 seconds earlier that it would have done if the rope had not broken.

Tkankyou

Thai
(b) is wrong. you've missed out the friction force in your calculation.
Reply 2
t - 400 = 800a
t = 800a + 400

2400-600-800a - 400 = 1200a
1400 = 2000a
a = 1400/2000 = 7/10 ms-2 (a)

t = 560 + 400 = 960N (b)

u = 20
v = 28
a = (2400 - 600)/1200 = 18/12 = 3/2
t = ?

t = 28-20 / (3/2) = 16/3 = 5 1/3

u = 20
v = 28
a = 7/10
t = 28-20 / (7/10) = 80/7 = 11 3/7

time diffrence = 80/7 - 16/3 = 240 - 112 / 21 = 128/21 = 6.095
For (c) what is the new acceleration, now that the truck is travelling on its own?
Once you know a, you are told u=20 v=28 use equations of motion to find how long it takes to reach 28 m/s.

Then do the same using acceleration of 0.7m/s had the rope not broken, and find how long it takes to get to 28 m/s. compare the times.
Codefusion, you just did all the work for him. great
Reply 5
Thankyou codefusion!
i was being sarcastic btw