A force of 4 N acts on a body of mass 2kg for 4secs. Assuming the body to be initially at rest, find the distance covered in 10 secs after the force starts acting.
Answers
Answer :
- Distance traveled by the body in 10s is 64 m.
Explanation :
Given :
- Force acting on the body, F = 4 N.
- Mass of the body, m = 2 kg.
- Time taken for the application of force, t = 4 s.
- Initial velocity, u = 0 m/s (since, the body is assumed to be in the state of rest).
To find :
- Distance covered by the body in the first 10 s = ?
Knowledge required :
- Formula for force exerted by the body, F = ma.
[Where : F, m and a are the force exerted, mass of the body and acceleration, respectively.]
- First equation of motion, v = u ± at.
[Where : v, u, a and t are the final velocity, initial velocity, acceleration and time taken, respectively.]
- Second equation of motion, S = ut ± ½at².
[Where : S, u, a and t are the distance covered, initial velocity, acceleration and time taken, respectively.]
- Formula for velocity of a body, v = S/t.
[Where : v, t and S are the velocity, time taken and distance covered, respectively.]
Solution :
First let us find the acceleration of the body of (mass 2 kg) by applying a force of 4 N :
By using the formula for force exerted by a body and substituting the values in it, we get :
⠀⠀⠀⠀⠀⠀=> F = ma
⠀⠀⠀⠀⠀⠀=> 4 = 2(a)
⠀⠀⠀⠀⠀⠀=> a = 2 m/s²
∴ Hence, the acceleration of the body is 2 m/s².
Now, we know that the initial velocity of the body is 0 m/s. So by using the second equation of motion, we can find the distance covered by the body in the 4 s. (Time of application of force).
By using the second equation of motion, we get :
⠀⠀⠀⠀⠀⠀=> S = ut + ½at²
⠀⠀⠀⠀⠀⠀=> S = 0(4) + ½(2)(4)²
⠀⠀⠀⠀⠀⠀=> S = 16 m
∴ Hence, the distance covered by the body in 4 s is 16 m.
Now, let us find the velocity attained by the body in 4 s. (Which will be uniform after 4 s).
By using the first equation of motion, we get :
⠀⠀⠀⠀⠀⠀=> v = u + at
⠀⠀⠀⠀⠀⠀=> v = 0 + 2(4)
⠀⠀⠀⠀⠀⠀=> v = 8 m/s
∴ Hence the velocity attained by the body is 8 m/s².
Now we have to find the distance covered by the body in 6 s, not under the application of force. [Distance covered in the 4 s (under the application of force is 16 m).
By using the formula for below and substituting the values in it, we get :
⠀⠀⠀⠀⠀⠀=> v = S/t
⠀⠀⠀⠀⠀⠀=> 8 = S/6
⠀⠀⠀⠀⠀⠀=> S = 48 m
∴ Hence, the distance covered by the body in 4 s is 48 m.
Now, the distance covered in 10 s = Distance traveled in 4s + Distance traveled in 6s :
⠀⠀⠀⠀⠀⠀=> S = 16 + 48
⠀⠀⠀⠀⠀⠀=> S = 64 m
∴ Hence, the distance covered by the body in 10 s is 64 m.
Answer:
Given :-
A force of 4 N acts on a body of mass 2kg for 4secs. Assuming the body to be initially at rest
To Find :-
find the distance covered in 10 secs after the force starts acting.
Solution :-
At first
A = f/m
A = 4/2
A = 2 m/s²
Now,
s = ut + ½at²
Where
s = distance travelled
u = initial velocity
t = time
a = acceleration
s = 0(4) + ½ × 2 × (4)²
s = 0 + 1 × 16
s = 0 + 16
s = 16 m
Now
v = u + at
Where
v = final velocity
u = initial velocity
a = acceleration
t = time
v = 0 + 2(4)
v = 0 + 8
v = 8 m/s
Now,
S = vt
S = (8)(4)
S = 32 m
Total distance covered = 16 + 48 = 64 m