Question

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters and t in seconds. At t = 1.1 s, what are (a) the magnitude and (b) the angle (within (-180°, 180°] interval relative to the positive direction of the x axis) of the net force on the particle, and (c) what is the angle of the particle's direction of travel?

Answer 1

Part a)

`F = 7.76 N`

Part b)

`\theta = -137.7 degree`

Part c)

`\theta = -127.7 degree`

Step-by-step explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

`x = -19 + t - 3t^3`

`y = 20 + 7t - 9t^2`

Part a)

differentiate x and y two times with respect to time to find the acceleration

`a_x = (d^2)/(dt^2)(-19 + t - 3t^3)`

`a_x = (d)/(dt)(0 +1 - 9t^2)`

`a_x = -18t`

`a_y = (d^2)/(dt^2)(20 + 7t - 9t^2)`

`a_y = (d)/(dt)(0 +7 - 18t)`

`a_y = -18`

Now the acceleration of the object is given as

`\vec a = (-18t)\hat i + (-18)\hat j`

at t= 1.1 s we have

`\vec a = -19.8 \hat i - 18 \hat j`

now the net force of the object is given as

`\vec F = m\vec a`

`\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)`

`\vec F = -5.74 \hat i - 5.22 \hat j`

now magnitude of the force will be

`F = √(5.74^2 + 5.22^2) = 7.76 N`

Part b)

Direction of the force is given as

`tan\theta = (F_y)/(F_x)`

`tan\theta = (-5.22)/(-5.74)`

`\theta = -137.7 degree`

Part c)

For velocity of the particle we have

`v_x = \frac{dx}[dt}`

`v_x = (0 +1 - 9t^2)`

`v_y = (dy)/(dt)`

`v_y = (0 +7 - 18t)`

now at t = 1.1 s

`\vec v = -9.89\hat i - 12.8 \hat j`

now the direction of the velocity is given as

`\theta = tan^(-1)((v_y)/(v_x))`

`\theta = tan^(-1)((-12.8)/(-9.89))`

`\theta = -127.7 degree`