Question

A car starts from rest and gains the velocity of

30m/s in 10s . If the mass of the car is 400kg.What

is the force applied on the car ?​

Answer 1

Answer:

20s and force is8newton

Answer 2

Given :

• A car starts from rest and gains velocity of 30 m/s in 10s
• Mass of the car = 400 kg

To Find :

• The Force applied on the car

Solution :

Using the First equation of motion ,

$\\ \star \: {\boxed{\sf{\purple{ \: v = u + at \: }}}}$

$\\ {\sf{where}} \begin{cases}& {\sf{v \: is \: final \: velocity}} \\ & {\sf{u \: is \: initial \: velocity}} \\ &{\sf{t \: is \: time}}\\ &{\sf{a \: is \: acceleration}}\end{cases}$

⠀⠀━━━━━━━━━━━━━━━━━━━━━━━

$\\ {\sf{We\:have}} \begin{cases}& \sf{v\:=30\:ms^{-1}}\\ & \sf{t\:=\:10\:s}\\ & \sf{m\:=\:400\:kg}\\ &\sf{u\:=\:0\:(starting\:from\:rest)}\end{cases}$

Substituting the values ,

$\\ : \implies \sf \: 30 = 0 + a(10)$

• Since a = $\sf{\dfrac{F}{m}}$

$\\ : \implies \sf \: 30 = 0 + \bigg( \frac{F}{m} \bigg)(10)$

$\\ : \implies \sf \: 30 = 0 + \bigg( \frac{F}{400} \bigg)(10)$

$\\ : \implies \sf \: 30 = \frac{F}{40}$

$\\ : \implies \sf \: F = 40 \times 30$

$\\ : \implies{\underline{\boxed{\pink{\mathfrak{ \: F = 1200 \: N}}}}} \bigstar$

$\\ \therefore {\underline{\sf{Hence \: , \: The \: Force \: applied \: on \: the \: car \: is \: \bold{1200 \: N}}}}$