Question

Given mass is 2*10^3kg speed is 36 km per hour distance is 50 m calculate average force to stop and time to stop it

Answer 1

Explanation:

u = 36 km/hr = 10 m/s

v = 0 m/s

S = 50 m

$a = \frac{ {v}^{2} - {u}^{2} }{2s} = \frac{100 - 0}{2 \times 50} = 1 \: m {s}^{ - 2}$

Now ,

$force = mass \times acceleration$

$force = 2000 \times 1 = 2000 \: n$

Hence , the average force required to stop is 2000 N.

I hope it helps you. If you have any doubts, then don't hesitate to ask.

Answer 2

Answer:

• Force required to stop the body = -2000 N

[ Negative sign shows that direction of force required is opposite to the motion of body ]

• time taken to stop the body = 10 sec

Explanation:

Given

• Mass of body, m = 2 × 10³ kg
• Initial velocity/speed, u = 36 km/h = 36 × 5/18  m/s = 10 m/s
• Final velocity of body, v = 0    [since, it is brought to rest]
• Distance covered, s = 50 m

To find

• Average force to stop the body, F =?
• Average time to stop it, t =?

Formulae required

• Third equation of motion

         2 a s = v² - u²

• First equation of motion

         v = u + a t

• Newton's second law of motion

        F = m a

[ Where v is final velocity, u is initial velocity, a is acceleration, s is distance covered, m is mass, t is time taken and F is force required ]

Solution

Let, acceleration of body be a

then,

Using third equation of motion

→ 2 a s = v² - u²

→ 2 a (50) = (0)² - (10)²

→ 100 a = -100

→ a = -1 m/s²

Now,

Using expression for Newton's second law of motion

→ F = m a

→ F = ( 2 × 10³ ) · ( -1 )

→ F = -2000 N

Further,

Using first equation of motion

→ v = u + a t

→ 0 = ( 10 ) + ( -1 ) t

→ -10 = -t

→ t = 10 sec

Therefore,

• Force required to stop the body is 2000 N in the direction opposite to the motion of body.
• Time taken to stop the body is 10 sec.