Question

A truck starts from rest and roll down a a hill with constant acceleration. It travels a distance of 400 metre in 20 sec. Find its acceleration. Find the force acting on it if its mass is 7000 kg

Answer 1

Question :–

A truck starts from rest and roll down a a hill with constant acceleration. It travels a distance of 400 metre in 20 sec. Find its acceleration. Find the force acting on it if its mass is 7000 kg.

Answer :–

1. Acceleration = 2 ${m/s}^{2}$.
2. Force acting = 14,000N.

Given :–

• Initial velocity (u) = 0 [starts from rest]
• Distance travelled (s) = 400 m
• Time taken (t) = 20 s
• Mass = 7000 kg

To Find :–

1. Find its acceleration (a)
2. Find the force acting on it if its mass is 7000 kg.

Solution :–

1. Using the equation of motion,

S = ut + $\frac{1}{2} a \times {t}^{2}$

where, u = 0

So, the equation will be,

S = $\frac{1}{2} a \times {t}^{2}$

a = $\frac{2 \times S}{{t}^{2}}$

→ $\frac{2 \times 400}{{20}^{2}}$

→ $\frac{800}{20 \times 20}$

→ $\frac{8 \cancel{00}}{4 \cancel{00}}$

→ $\frac{\cancel8}{\cancel4}$

→ 2 ${m/s}^{2}$

Hence,

The acceleration = 2 ${m/s}^{2}$

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2. Using newton's 2nd law of motion,

Force = Mass × Acceleration

→ 7,000 × 2

→ 14,000N

Hence,

The force acting on it if its mass is 7,000 kg is 14,000N.

Answer 2

Answer:

Given :

Mass of the bus = 7000 kg

Time = 4 sec

Velocity of the bus = 36 km/hr

To Find :

The Force applied to stop the bus

Solution :

using first equation of motion , v = u + at

Here ,

v is final velocity

u is initial velocity

a is acceleration

t is time

We have ,

v = 0 {since it comes to rest}

u = 36 km/hr = 36(5/18) m/s = 10 m/s

t = 4 sec

Mass = 7000 kg

Substituting the values ;

➙ 0 = 10 + (F/m)(4) { ∵ a = F/m}

➙ 0 = 10 + (F/7000)(4)

➙ 0 = 10 + (4F/7000)

➙ 0 = (70000 + 4F)/7000

➙ 0 = 70000 + 4F

➙ -70000 = 4F

➙ F = -70000/4

➙ F = -17500 N \begin{gathered}\\\end{gathered}

Hence ,

The Force that should be applied to bring the bus to rest is -17,500 N