Question

A car of mass 2.5 ton travelling at a velocity of 36 km/h increases its velocity to 72 km/h in 15 seconds. Calculate the force exterted by the car.

Answer 1

Answer:

Force exerted = 125000/3 N = 41666.67 N

Explanation:

We know that,

Force = Rate of change of momentum

Force = Δp/Δt

Given that,

Mass = 2.5 ton = 25000 kg

Initial velocity = u = 36km/hr = 36 * 5/18 = 10m/s

Final velocity = v = 72 km/hr = 72 * 5/18 = 20m/s

Hence

Initial momentum = mu = 25000 * 10 = 250000 kg m/s

Final momentum = mv = 25000 * 20 = 500000 kg m/s

Time taken = 15s

Therefore,

Δp = mv - mu = 500000 - 250000 = 250000

Δt = 15s

Hence

Force = Δp/Δt

Force = 250000/15

Force = 125000/3 N = 41666.67 N

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Force\:exerted=\frac{5000}{3}\:N}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Mass \: of \: car (m)= 2.5 \: ton \\ \\ \tt: \implies Initial \: velocity(u) = 36 \: km/h \\ \\ \tt: \implies Final \: velocity(u) = 72\: km/h \\ \\ \tt: \implies Time(t) = 15 \:sec \\ \\ \red{\underline \bold{To \: find :}} \\ \tt: \implies Force(F) =?$

• According to given question :

$\tt \circ \: Initial \: velocity = 36 \times \frac{5}{18} = 10 \: m/s \\ \\ \tt \circ \: Final \: velocity = 72 \times \frac{5}{18} = 20 \: m/s\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 20 = 10 + a \times 15 \\ \\ \tt: \implies 20 - 10 = a \times 15 \\ \\ \tt: \implies a = \frac{10}{15} \\ \\ \green{\tt: \implies a = \frac{2}{3} \: m/{s}^{2} } \\ \\ \tt \circ \: Mass = 2.5 \times 1000 = 2500 \: kg \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies F = ma \\ \\ \tt: \implies F=2500 \times \frac{2}{3} \\ \\ \green{\tt: \implies F= \frac{5000}{3} \: N} \\ \\ \green{\tt \therefore Force \: exerted \: by \: car \: is \: \frac{5000}{3} \: N}$