
A body of mass 10 kg is moving eastward with a uniform speed of 2 m/s. A force of 20 N is applied to it toward the north. What is the magnitude of displacement after 2 seconds?
A. $4~m$
B. $8~m$
C.$ 4~ \sqrt {2}$
D. $8~\sqrt {2}$
Answer
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Hint: Since the force is applied on a body in north direction which is moving eastward, there is a change in speed as well as direction of a moving body. Also, the direction of motion of a body and the force applied is perpendicular to each other so we will apply second equation of motion to solve the given problem.
Complete answer:
Speed of the body towards east = $2m/s$(given) = ${u_x}$
No acceleration in the direction of motion as body is moving with uniform speed, ${a_x} = 0$
Mass of the body = $10kg$(given)= $m$
Force Applied on the body towards north = $20N$(given)= $F$
Time Taken = $2s$ (given) = $t$
Since, the force is applied towards the North direction, acceleration i.e., change in velocity will take place in the direction of force applied. Hence, By Applying Newton’s Second Law of motion:
$acceleration = \dfrac{{Force(F)}}{{mass(m)}}$
$a = \dfrac{{20}}{{10}} = 2m/{s^2}$(towards North) = ${a_y}$. The obtained values can be written as:
Now, we know from second equation of motion, distance travelled is:
$s = ut + \dfrac{1}{2}a{t^2}$
Applying this equation along x-axis:
${s_x} = {u_x}t + \dfrac{1}{2}{a_x}{t^2}$
${s_x} = 2 \times 2 + \dfrac{1}{2} \times 0 \times {2^2} = 4m$
Along y-axis:
${s_y} = {u_y}t + \dfrac{1}{2}{a_y}{t^2}$
${s_y} = 0 \times 2 + \dfrac{1}{2} \times 2 \times {2^2} = 4m$
Now, Magnitude of displacement after $2s$ can be calculated as:
$d = \sqrt {{s_x}^2 + {s_y}^2} $
\[d = \sqrt {{4^2} + {4^2}} = 4\sqrt 2 m\]
Thus, the magnitude of displacement of a body after $2s$ is $4\sqrt 2 m$.
Hence, the correct option is: (C)$4\sqrt 2 m$
Note: Since this is a problem related to both equations of motion and laws of motion hence, given conditions is to be analysed carefully as on the basis of it only the procedure of solving the problem is identified. Figure must be drawn while visualising the problem as it gives better understanding of using the formulas. Units must be put after each end result.
Complete answer:
Speed of the body towards east = $2m/s$(given) = ${u_x}$
No acceleration in the direction of motion as body is moving with uniform speed, ${a_x} = 0$
Mass of the body = $10kg$(given)= $m$
Force Applied on the body towards north = $20N$(given)= $F$
Time Taken = $2s$ (given) = $t$
Since, the force is applied towards the North direction, acceleration i.e., change in velocity will take place in the direction of force applied. Hence, By Applying Newton’s Second Law of motion:
$acceleration = \dfrac{{Force(F)}}{{mass(m)}}$
$a = \dfrac{{20}}{{10}} = 2m/{s^2}$(towards North) = ${a_y}$. The obtained values can be written as:
| Along x-axis (direction of motion) | Along y-axis (direction of force applied) |
| \[{u_x} = 2m/s\] | \[{u_y} = 0\] |
| \[{a_x} = 0\] | \[{a_y} = 2m/{s^2}\] |
Now, we know from second equation of motion, distance travelled is:
$s = ut + \dfrac{1}{2}a{t^2}$
Applying this equation along x-axis:
${s_x} = {u_x}t + \dfrac{1}{2}{a_x}{t^2}$
${s_x} = 2 \times 2 + \dfrac{1}{2} \times 0 \times {2^2} = 4m$
Along y-axis:
${s_y} = {u_y}t + \dfrac{1}{2}{a_y}{t^2}$
${s_y} = 0 \times 2 + \dfrac{1}{2} \times 2 \times {2^2} = 4m$
Now, Magnitude of displacement after $2s$ can be calculated as:
$d = \sqrt {{s_x}^2 + {s_y}^2} $
\[d = \sqrt {{4^2} + {4^2}} = 4\sqrt 2 m\]
Thus, the magnitude of displacement of a body after $2s$ is $4\sqrt 2 m$.
Hence, the correct option is: (C)$4\sqrt 2 m$
Note: Since this is a problem related to both equations of motion and laws of motion hence, given conditions is to be analysed carefully as on the basis of it only the procedure of solving the problem is identified. Figure must be drawn while visualising the problem as it gives better understanding of using the formulas. Units must be put after each end result.
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