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Question: A time-dependent force, \( F = \left( {8.00i{\text{ }}-{\text{ }}4.00tj} \right)N \) , where \( t \)...

A time-dependent force, F=(8.00i  4.00tj)NF = \left( {8.00i{\text{ }}-{\text{ }}4.00tj} \right)N , where tt is in seconds, is exerted on a 2.00kg2.00kg object initially at rest.
A) At what time will the object be moving with a speed of 15.0 m/s15.0{\text{ }}m/s ?
B) How far is the object from its initial position when its speed is 15.0 m/s15.0{\text{ }}m/s ?
C) Through what total displacement has the object travelled at this moment?

Explanation

Solution

For part A) first find the acceleration using Newton's second law. Integrate it with respect to time. We will get a relation between velocity and time. Find the value of time. For part B) use the formula of distance to find the distance of the object from its initial position. for part C) find the displacement by the difference between final and initial position of the object.

Complete Step By Step Answer:
It is given that the F=(8.00i  4.00tj)NF = \left( {8.00i{\text{ }}-{\text{ }}4.00tj} \right)N mass of the object is 2.00kg2.00kg and speed of the object is 15.0 m/s15.0{\text{ }}m/s

Solution of part A)
Using Newton's second law of motion.
F=maF = ma
a=Fm\Rightarrow a = \dfrac{F}{m}
a=8.00i+(4.00t)j2\Rightarrow a = \dfrac{{8.00i + (4.00t)j}}{2}
a=4.00i+(2.00t)j\Rightarrow a = 4.00i + (2.00t)j
Integrating the above equation
dvdt=4.00i+(2.00t)j\dfrac{{dv}}{{dt}} = 4.00i + (2.00t)j
v=(4.00t)i+(t2)j\Rightarrow v = (4.00t)i + ({t^2})j
Putting the value of speed
15=(4.00t)i+(t2)j\Rightarrow 15 = (4.00t)i + ({t^2})j
152=(4.00t)2+(t2)2\Rightarrow {15^2} = {(4.00t)^2} + {({t^2})^2}
t4+16t2=225\Rightarrow {t^4} + 16{t^2} = 225
t2=9sec\Rightarrow {t^2} = 9\sec
t=3sec\Rightarrow t = 3\sec
Hence, the time in which the object will be moving with a speed of 15.0 m/s15.0{\text{ }}m/s is 3sec3\sec

Solution for part B)
Using the distance formula speed is equal to distance divided by time
dxdt=(4.00t)i+(t2)j\dfrac{{dx}}{{dt}} = (4.00t)i + ({t^2})j
On integrating the above equation
x=(2.00t2)i+t33jx = (2.00{t^2})i + \dfrac{{{t^3}}}{3}j
Putting the value of time from above,
x=(2×32)+(3)33\left| x \right| = \sqrt {(2 \times {3^2}) + \dfrac{{(3){}^3}}{3}}
x=5.19m\Rightarrow x = 5.19m
Hence, the object is distance from its initial position when its speed is 15.0 m/s15.0{\text{ }}m/s is 5.19m5.19m

Solution for part C)
the total displacement the object has travelled at this moment in vector form is
d=18i9jd = 18i - 9j .

Note:
The three Newtonian laws of motion give the relation between an object's motion and the forces acting on it. According to Newton's second law, the rate of change of momentum is proportional to the force applied for the given object, or that the force on an object is equal to the product of mass and the acceleration.