Question

A 1200 kg car pushes a 2200 kg truck that has a dead battery . When the driver stes on he accelerator, the drive wheels of the car push against the ground with force of 4000N. Rolling friction ca be neglected.

(a) What is the magnitude of the force of the car on the truck?
Express you answer to two significant figures and include the appropriate units.
(b) What is the magitude of the force of the truck on the car?
Express your answer to two significant figures and include the appropriate units.

Answer 1

Answer:

Part (i)  the magnitude of the force of the car on the truck ($F_c$) is 1400 N

Part(ii) the magnitude of the force of the truck on the car ($F_T$) is 2600 N

Explanation:

Let  the mass of the car = $M_c$

Let the mass of the truck with dead battery =$M_T$

When the car pushes the truck, they accelerate  with a (m/s²)

Total force on both the car and the truck, F = 4000N

Given;

$M_c$ = 1200kg

$M_T$ = 2200kg

Substitute in these values in the equation below and calculate their common acceleration.

$F = (M_c+M_T)a\\\\4000 =(1200+2200)a\\\\4000 = (3400)a\\\\a =\frac{4000}{3400} = 1.1765\frac{m}{s^2}$

Part (i) the magnitude of the force of the car on the truck ($F_c$)

$F_c$ = $M_c$ × a

   = 1200 × 1.1765 = 1400 N (in two significant figures)

Part(ii) the magnitude of the force of the truck on the car ($F_T$)

$F_T$ = $M_T$ × a

     =  2200 × 1.1765 = 2600 N  (in two significant figures)

Answer 2

The magnitude of the force of the car on the truck and the magnitude of the force of the truck on the car is mathematically given as

a)Fc = 1400 N

b)Ft = 2600 N

Magnitude of the force

Question Parameters:

A 1200 kg car pushes a 2200 kg truck that has a dead battery the drive wheels of the car push against the ground with force of 4000N

Generally the equation for the Force   is mathematically given as

F=(Mc+Mt)a

Therefore

4000=(1200+200)a

a=1.176m/s

Therefore

a)

Fc = M_c × a

Fc=  = 1200 × 1.1765

Fc = 1400 N

b)

Ft = M_T × a

Ft=  2200 × 1.1765

Ft = 2600 N

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