Question

An object (m=0.5kg) is moving along x-axis with v0 = 1.6m/s. At t =0s, at which object is at thr origin, a force directed along the negative x-axis F = -kt^3 starts to apply to it. At t=2s, the object passes the origin again. Find k ( a constant).

Answer 1

We can use the relation between force and acceleration as:

$F=m{a}=m\cfrac{dv}{dt} \implies dv=\cfrac{F}{m}dt$

Then, we can integrate by using the limits v_i=1.6 m/s at t_i = 0s and we will consider that v_f= 0 m/s when t_f = 2 s:

$\int_{v_i}^{v_f} dv=\int_{t_i}^{t_f} \cfrac{F}{m}\,dt=\int_{t_i}^{t_f} \cfrac{-kt^3}{m}\,dt \\ {v_f}-{v_i} =-\cfrac{k}{m}\int_{t_i}^{t_f} t^3\,dt=-\cfrac{k}{4m}[t^4_f-t^4_i] \\ \text{ } \\ \implies k=\dfrac{4m({v_i}-{v_f})}{t^4_f-t^4_i}$

After substitution, we can go ahead and find the value for k:

$k=\dfrac{4(0.5\,kg)({1.6\frac{m}{s}}-0\frac{m}{s})}{(2\,s)^4-(0\,s)^4} \\ k=\dfrac{(2)({1.6\frac{kg \cdot m}{s}})}{16\,s^4} \\\implies k=0.2\cfrac{kg \cdot m}{s^5}=0.2\cfrac{N}{s^3}$

In conclusion, we find that k = 0.2 N/s³.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.