Question

The force F on an object is the product of the mass m and the acceleration a. In this problem, assume that the mass and acceleration both depend on time t, hence so the does the force. That is, F(t)=m(t)a(t) At time t=7 seconds, the mass of an object is 49 grams and changing at a rate of −1gs. At this same time, the acceleration is 11ms2 and changing at a rate of −10ms3. By the product rule, the force on the object is changing at the rate of .

Answer 1

`(dF)/(dt) = -0.501\,(N)/(s)`

Step-by-step explanation:

The rate of change of the force is:

`(dF)/(dt) = (dm)/(dt)\cdot a(t) + m(t) \cdot (da)/(dt)`

`(dF)/(dt) = (-1* 10^(-3)\,(kg)/(s) )\cdot (11\,(m)/(s^(2))) + (0.049\,kg)\cdot (-10\,(m)/(s^(3)) )`

`(dF)/(dt) = -0.501\,(N)/(s)`

Answer 2

Answer:
`-501* 10^(-3) N/s`

Step-by-step explanation:

Given

`F\left ( t\right )=m\left ( t\right )a\left ( t\right )`

at t=7 s

m=49 gm

`\frac{\mathrm{d} m}{\mathrm{d} t}=-1 gm/s`

`a=11 m/s^2`

`\frac{\mathrm{d} a}{\mathrm{d} t}=-10 m/s^3`

Differentiating Force

`\frac{\mathrm{d} F(t)}{\mathrm{d} t}=\frac{\mathrm{d} m}{\mathrm{d} t}a+\frac{\mathrm{d} a}{\mathrm{d} t}m`

`\frac{\mathrm{d} F(t)}{\mathrm{d} t}=-1* 11-10* 49=-11-490=-501* 10^(-3) N/s`