Given: Mass m = 18 Kg; Height h = 3.6 m; Time t = 60 s

Required: Power output in unit of Watts

Formula: P = mgh/t

P = (18 Kg)(9.8 m/s²)(3.6 m)/60 s

P = 10.58 J/s or Watts

7 0

3 years ago

How do you solve this problem?

Katarina [22]

The particle has acceleration vector

$\vec a=\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\jmath$

We're told that it starts off at the origin, so that its position vector at $t=0$ is

$\vec r_0=\vec0$

and that it has an initial velocity of 12 m/s in the positive $x$ direction, or equivalently its initial velocity vector is

$\vec v_0=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\,\vec\imath$

To find the velocity vector for the particle at time $t$ , we integrate the acceleration vector:

$\vec v=\vec v_0+\displaystyle\int_0^t\vec a\,\mathrm d\tau$

$\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath$

$\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath$

Then we integrate this to find the position vector at time $t$ :

$\vec r=\vec r_0+\displaystyle\int_0^t\vec v\,\mathrm d\tau$

$\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath$

$\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath$

Solve for the time when the $y$ coordinate is 18 m:

$18\,\mathrm m=\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\implies t=3.0\,\mathrm s$

At this point, the $x$ coordinate is

$\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)(3.0\,\mathrm s)+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3.0\,\mathrm s)^2=27\,\mathrm m$

so the answer is C.

7 0

3 years ago

Distance is constant and time increseas Will Speed increase or decrease?​

Distance is constant and time increseas Will Speed increase or decrease?