A force acts on a trolley of mass 50 kg for 8 s. It moves through a distance of 30 m in the next 3 s. Calculate the :
(a) velocity attained
(b) acceleration produced, and
(c) magnitude of force applied
Answers
Answered by
15
As we know :
V =
V =
V = 10m/s
b)Now,
v = u + at
=> 10 = 0 + a × 8
=> 10 = 8a
=> a = 10/8 = 1.25 m/s^2
c)
F = ma
F= 50×1.25
F = 62.5 N
What is force ?
=> A pull or push on an object is called force. It can start ,stop a body or change its direction of motion.
"N" (Newton) is the SI unit of Force .
In CGS it's Unit is Dyne .
1 Newton =
Velocity :
It is a vector quantity and it is defined as the ratio of displacement to the total time. Velocity maybe zero or positive or negative.
Velocity =
Acceleration :
It is defined as the rate of change of velocity.
Acceleration =
#Be Brainly !!
V =
V =
V = 10m/s
b)Now,
v = u + at
=> 10 = 0 + a × 8
=> 10 = 8a
=> a = 10/8 = 1.25 m/s^2
c)
F = ma
F= 50×1.25
F = 62.5 N
What is force ?
=> A pull or push on an object is called force. It can start ,stop a body or change its direction of motion.
"N" (Newton) is the SI unit of Force .
In CGS it's Unit is Dyne .
1 Newton =
Velocity :
It is a vector quantity and it is defined as the ratio of displacement to the total time. Velocity maybe zero or positive or negative.
Velocity =
Acceleration :
It is defined as the rate of change of velocity.
Acceleration =
#Be Brainly !!
Answered by
12
Let's understand the question first.
A force acts on a trolley of mass 50 kg for 8s.
So Mass of trolley is 50kg. initially it was at rest. Then an unknown force(let's say F) acts on it for 8 seconds. This force will increase in velocity from 0 to some v in 8 seconds.
It moves through a distance of 30 m in the next 3 s.
So force is removed at 8th second. Now the trolley has attained velocity v and moves at constant velocity v. Within 3 second, it moved 30m.
So from t=0 to t=8 sec, the trolley was in an accelerated motion.
from t=8s to t=11s, it was moving at constant velocity v.
Coming to the questions:
a.
We need to find the velocity attained.
I have explained the velocity attained is v.
It had moved 30m in 3s at constant velocity v.
we know displacement = v × t
=> 30 = v × 3
=> v = 30/3
=> v = 10 m/s
(b)
We need to find the acceleration produced
acceleration was produced in the first 8 second of the motion when force was acting.
During these 8 sec, velocity changed from 0 to v(=10 m/s)
Use v = u + at
=> 10 = 0 + a × 8
=> 10 = 8a
=> a = 10/8 = 1.25 m/s^2
acceleration produced is 1.25 m/s^2.
(c)
We need to find the magnitude of force applied
This is straight forward. We know F = ma
=> F = 50 × 1.25
=> F = 62.5 N
Hence, magnitude of force applied is 62.5 N
Any doubt, ask in comment.
A force acts on a trolley of mass 50 kg for 8s.
So Mass of trolley is 50kg. initially it was at rest. Then an unknown force(let's say F) acts on it for 8 seconds. This force will increase in velocity from 0 to some v in 8 seconds.
It moves through a distance of 30 m in the next 3 s.
So force is removed at 8th second. Now the trolley has attained velocity v and moves at constant velocity v. Within 3 second, it moved 30m.
So from t=0 to t=8 sec, the trolley was in an accelerated motion.
from t=8s to t=11s, it was moving at constant velocity v.
Coming to the questions:
a.
We need to find the velocity attained.
I have explained the velocity attained is v.
It had moved 30m in 3s at constant velocity v.
we know displacement = v × t
=> 30 = v × 3
=> v = 30/3
=> v = 10 m/s
(b)
We need to find the acceleration produced
acceleration was produced in the first 8 second of the motion when force was acting.
During these 8 sec, velocity changed from 0 to v(=10 m/s)
Use v = u + at
=> 10 = 0 + a × 8
=> 10 = 8a
=> a = 10/8 = 1.25 m/s^2
acceleration produced is 1.25 m/s^2.
(c)
We need to find the magnitude of force applied
This is straight forward. We know F = ma
=> F = 50 × 1.25
=> F = 62.5 N
Hence, magnitude of force applied is 62.5 N
Any doubt, ask in comment.
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