Question

A truck starts from rest and rolls down the hill with a constant acceleration. It travels 400 m in 20 s. If the mass of truck is 7000 Kg, the force acting on it is:​

Answer 1

Answer

Given -

• $\rm u = 0$ ( because it starts from rest )
• $\rm s = 400 m$
• $\rm t = 20 sec$
• $\rm m = 7000 kg$

where

$\longrightarrow$u is initial velocity.

$\longrightarrow$s is distance travelled.

$\longrightarrow$t is time taken.

$\longrightarrow$m is mass of truck.

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To find -

Force acting $\longrightarrow$ F

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Formula used -

$\boxed{\bf\red{s = ut + \frac{1}{2}at^2}}$

$\boxed{\bf\pink{ F = m \times a} }$

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Solution -

For calculating force we first need to find acceleration.

Calculating Acceleration -

• $\rm u = 0$
• $\rm s = 400 m$
• $\rm t = 20 sec$

Substituting the value in 2nd equation of motion -

$\implies$$\rm s = ut + \frac{1}{2}at^2$

$\implies$$\rm 400 = 0 + \frac{1}{2} \times a \times {20}^2$

$\implies$$\rm 400 = 200a$

$\implies$$\rm a = \frac{\cancel {400}}{\cancel {200}}$

$\implies$$\rm a = 2 m/s^2$

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Calculating Force -

• $\rm m = 7000 kg$
• $\rm a = 2 m/s^2$

Substituting the value in formula -

$\implies$$\rm F = ma$

$\implies$$\rm F = 7000 \times 2$

$\implies$$\rm F = 14000 N$

$\boxed{\underline{\rm\red{Force \:acting \:on \:it\: is \:14000 N.}}}$

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Answer 2

AnsWer :

• Force acting on the truck is 14, 000 N

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Given :

• Initial Velocity = 0 ( starts from rest )

• Distance = 400 m

• Time = 20 s

• Mass of the truck = 7000 kg

To Find :

• Force acting on the truck

Solution :

☯ First , let us find acceleration .

$\small{\boxed{\boxed{\sf{\red{s\:=\:ut\:+\: \dfrac{1}{2}at²}}}}}$

$\implies \small \sf{400\:=\:0(20)\:+\: \dfrac{1}{2}a(20)²}$

$\implies \small \sf{400\:=\:0\:+\: \dfrac{1}{2}a(400)}$

$\implies \small \sf{400\:=\:\dfrac{400a}{2}}$

$\implies \small \sf{400\:=\:200a}$

$\implies \small \sf{a\:=\:\dfrac{400}{200}}$

$\implies \large \sf{\red{Acceleration\:=\:2\:m/s²}}$

☯ Now , let us find the Force on the truck .

$\small{\boxed{\boxed{\sf{\red{F\:=\:ma}}}}}$

$\implies \small \sf{Force\:=\:7000(2)}$

$\implies \large \sf{\red{Force\:=\:14, 000\:N}}$

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