Question

a 5 kg cart is moving horizontally at6 m/ s in order to change its speed to10 m/s the work done oncart
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Answer 1

Answer:

A 5 kg object is moving horizontally at 6 m/s. In order to change its speed to 10 m/s, the net work done on the object must be?

Data: Initial velocity, u=6m/s; Final velocity, v=10 m/s ; Mass of the object is ,

m=5 kg

The basic equation of the work done is W=F×d where F is the force acting

on the object and d is the distance covered.

The equation W=Fd is not suitable for the data, it is made suitable as following,

W=F×d

W=m×a×d ; ∵ F=ma

We know that, v2−u2=2ad

In this case u=0 ∵ work done is calculated from the rest , then the above equation

will be

ad=v22

⇒Work done from rest is, W=12mv2

The work done by the object to attain the velocity 6m/s is

W6=125kg×(6m/s)2

W6=90J

Similarly, W10=125kg×(10m/s)2=250J

In order to change the velocity from 6m/s to 10 m/s the net work done is

W10−W6=250−90=160J

Accordingly, work done by an object from rest

Answer 2

Answer:

160J

Explanation:

vi=6m/s             vf=10m/s           m=5kg

W=Fd

as  F=ma    so; W=mad

from 3 equation of motion we will find acceleration.

$vf^{2} -vi^{2} =2ad$

$10^{2} -6^{2} =2ad$

$ad=32$

so; W=(5)(32)

W=160J