Answer :
Answer:
[tex]3MQRT^2+\frac{9}{2}MR^2T^4[/tex]
Explanation:
In order to find the work done by the force, we use the work-energy theorem, which states that the work done by a force on an object is equal to the change in kinetic energy of the object. Mathematically:
[tex]W=K_f-K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex] (1)
where
W is the work done
m is the mass of the object
v is the final speed of the object
u is the initial speed
In this problem, we have:
m = M is the mass of the object
The position of the object at time t is
[tex]x(t)=P+Qt+Rt^3[/tex]
We can find its speed at time t by calculating the derivative of the position:
[tex]v(t)=x'(t)=Q+3Rt^2[/tex]
Therefore:
- The speed at time t = 0 is
[tex]u=v(0)=Q[/tex]
- The speed at time t = T is
[tex]v=v(T)=Q+3RT^2[/tex]
Substituting into eq.(1), we find the work done:
[tex]W=\frac{1}{2}M(Q+3RT^2)^2-\frac{1}{2}MQ^2=\\=\frac{1}{2}MQ^2+3QRT^2+\frac{9}{2}MR^2T^4-\frac{1}{2}MQ^2=\\=3MQRT^2+\frac{9}{2}MR^2T^4[/tex]