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

Answer

• Acceleration , a = 2 m/s²
• Force , F = 14000 N

Given

• A truck starts from rest and roll down a a hill with constant acceleration. It travels a distance of 400 metre in 20 sec

To Find

• Acceleration , a
• Force acting , F if it's mass is 7000 kg

Concept Used

Equation of motion

• s = ut + ¹/₂ at²

Newton's Second Law of Motion

• F = ma

Solution

Initial velocity , u = 0 m/s [ Starts from rest ]

Distance , s = 400 m

Time , t = 20 s

Acceleration , a = ? m/s²

Mass , m = 7000 kg

Force , F = ? N

Apply 3 rd equation of motion .

⇒ s = ut + ¹/₂ at²

⇒ 400 = (0)(20) + ¹/₂ (a)(20)²

⇒ 400 = 200a

⇒ 200a = 400

⇒ a = 2 m/s²

__________________________

Acceleration , a = 2 m/s²

__________________________

Apply Newton's Second Law .

⇒ F = ma

⇒ F = (7000)(2)

⇒ F = 14000 N

__________________________

Force , F = 14000 N

__________________________

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