An object of mass 2kg taken to a vertical height of 3 m. thereafter horizontally through a distance 2m and then vertically downward through 1m. total work done by gravity in whole process ( g = 10)
Answers
Given:
The mass of the body is m=2kg
The force applied on the body F=7N
The coefficient of kinetic friction =0.1
Since the body starts from rest, the initial velocity of body is zero.
Time at which the work is to be determined is t=10s
The acceleration produced in the body by the applied force is given by Newtons second law of motion as:
a′=mF=27=3.5m/s2
Frictional force is given as:
f=μg=0.1×2×9.8=1.96
The acceleration produced by the frictional force:
a"=−21.96=−0.98m/s2
Therefore, the total acceleration of the body:
a′+a"=3.5+(−0.98)=2.52m/s2
The distance traveled by the body is given by the equation of motion:
s=ut+21at2
=0+21×2.52×(10)2=126 m
(a) Work done by the applied force,
Wa=F⋅s=7×126=882 J
(b) Work done by the frictional force,
Wf=F⋅s=1.96×126=247 J
(c), (d)
From the first equation of motion, final velocity can be calculated as:
v=u+at
=0+2.52×10=25.
Given: An object of mass 2kg taken to height of 3 m, then 2 m horizontally and then vertically downward through 1 m. g= 10m/s^2
To find: Work done by gravity in whole process
Explanation: In first process, when object is taken upwards to a vertical height of 3 m , work done is negative since the displacement of body and direction of gravity is negative.
Work done in this process= mgh= - 2*10*3
= -60 J
In second process, the direction of gravity and displacement is in perpendicular direction. Therefore the work done is zero.
In third process, work done is positive since gravity and displacement is in same direction.
Work done in this process= 2*10*1
= 20 J
Therefore, the total work done= -60+20= -40 J
The total work done by gravity in this process is -40 J.