The displacement x of a body of mass 1 kg on horizontal smooth surface as a function of time t is given x= t 3 /3 . the work done by external agent in first one second is??
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x = t³ / 3 meters m = 10 kg
dx = t² dt
dx/dt = v = t²
d²x/dt² = a = 2 t
force acting on the body = m a = 20 t Newtons
![work\ done\ W= \int \limits_{t=0}^{t=t}\ {F} \ dx\\\\=\int \limits_{0}^t {20t * t^2}\ dt\\\\=\frac{20}{4}[t^4]_0^t=5\ t^4\\\\W(t=1sec)=5 Joules](https://tex.z-dn.net/?f=work%5C+done%5C+W%3D+%5Cint+%5Climits_%7Bt%3D0%7D%5E%7Bt%3Dt%7D%5C+%7BF%7D+%5C+dx%5C%5C%5C%5C%3D%5Cint+%5Climits_%7B0%7D%5Et+%7B20t+%2A+t%5E2%7D%5C+dt%5C%5C%5C%5C%3D%5Cfrac%7B20%7D%7B4%7D%5Bt%5E4%5D_0%5Et%3D5%5C+t%5E4%5C%5C%5C%5CW%28t%3D1sec%29%3D5+Joules "work\ done\ W= \int \limits_{t=0}^{t=t}\ {F} \ dx\\\\=\int \limits_{0}^t {20t * t^2}\ dt\\\\=\frac{20}{4}[t^4]_0^t=5\ t^4\\\\W(t=1sec)=5 Joules")
dx = t² dt
dx/dt = v = t²
d²x/dt² = a = 2 t
force acting on the body = m a = 20 t Newtons
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