A mass of 1 kg is moving from east to west with a velocity of 10 m/s. A force is applied to it for 2 seconds and its velocity becomes 5 m/s. What is the value and direction of the force?
Answers
Explanation:
Given,
Force act in direction of west F=10N
Velocity toward north Is V
n
=30 ms
−1
Initial velocity in west direction is u=0m/s
Acceleration in west direction is a=
M
F
=
1
10
=10m/s
2
Final velocity after time 4 sec, in west direction is V
s
=u+at=0+10×4=40m/s
Resultant velocity is V
R
=
V
s
2
+V
n
2
=
30
2
+40
2
=50ms
−1
Direction of resultant velocity makes angle θ with the direction of force act.
tanθ=
V
s
V
n
=
40
30
=
4
3
θ=tan
−1
(
4
3
)
Hence the velocity of the body after the force ceases will be 50 m/s todards an angle θ=tan
−1
(
4
3
) with the direction of force.
Answer:
The mass of body, m=0.40kg
Constant initial speed of the body is, u=10m/s
Force that acts on the body is, F=8.0N
(i)
At t= –5 s
The force starts acting on the body from t=0 s.
So, the acceleration of the body during this time was zero and it moves at a constant speed.
Position of the body is given by:
x=v×t
x=10×(−5)
=−50 m
This answer is percent correct