(a) For the work-energy theorem, the work done to lift the can of paint is equal to the gravitational potential energy gained by it, therefore it is equal to

$W=mg\Delta h$

where m=3.4 kg is the mass of the can, g=9.81 m/s^2 is the gravitational acceleration and $\Delta h=1.8 m$ is the variation of height. Substituting the numbers into the formula, we find

$W=(3.4 kg)(9.81 m/s^2)(1.8 m)=60.0 J$

(b) In this case, the work done is zero. In fact, we know from its definition that the work done on an object is equal to the product between the force applied F and the displacement:

$W=Fd$

However, in this case there is no displacement, so d=0 and W=0, therefore the work done to hold the can stationary is zero.

(c) In this case, the work done is negative, because the work to lower the can back to the ground is done by the force of gravity, which pushes downward. Its value is given by the same formula used in part (a):

$W=mg \Delta h=(3.4 kg)(9.81 m/s^2)(-1.8 m)=-60.0 J$

8 0

3 years ago

A turtle moves at an average speed of 0.2m/S towards the finish line that is 80m away from the starting line. A rabbit,racing wi

klasskru [66]

Answer:

5.6mls

Explanation:

3 0

3 years ago

A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

GrogVix [38]

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

$\frac{5}{6}mv^2 = mgh$

$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

$\frac{1}{2}mv^2 = mgh$

$h = \frac{v^2}{2g}$

$h = \frac{8^2}{2(9.81)} = 3.16 m$

3 0

3 years ago

4. show your understanding of electric and magnetic forces<br> below.

Vitek1552 [10]

Answer:

Explanation:

Electric forces exist among stationary electric charges; both electric and magnetic forces exist among moving electric charges. ... The magnetic force between two moving charges may be described as the effect exerted upon either charge by a magnetic field created by the other.

8 0

2 years ago