Question

A object has mass 5 kg and it is at rest. Now the object is pushed with a force of 10 N. How much distance will it cover in 5 mins?

Answer 1

Explanation:

u=0

t=300s

m=5kg

f=10

a=f/m=10/5=2m/s²

s=ut+1/2at²

s=0×300+1/2×2×90000

s=90000m

Answer 2

Given :

▪ Mass of object = 5kg

▪ Initial velocity = zero (i.e. rest)

▪ Applied force = 10N

▪ Time interval = 5min = 300s

To Find :

▪ Distance covered by object in the given interval of time.

Solution :

✒ First we have to find out acceleration of object (with the help of newton's second law of motion) after that we can calculate distance covered by object.

✒ Since, acceleration of object has said to be constant, we can easily apply equation of kinematics to solve this type of questions.

____________________________

$\bigstar\:\underline{\boxed{\bf{\red{F=m\times a}}}}\\ \\ \bigstar\: \underline{\boxed{\bf{\green{d=ut+\dfrac{1}{2}at^2}}}}$

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⏭ F = m × a

⏭ 10 = 5 × a

⏭ a = 10/5

⏭ a = 2m/s^2

→ d = ut + (1/2)at^2

→ d = (0×300) + [0.5×2×(300)^2]

→ d = (300)^2

→ d = 90000m = 90km