Question

A can of beans that has mass M is launched by a spring-powered device from level ground. The can is launched at an angle of α0 above the horizontal and is in the air for time T before it returns to the ground. Air resistance can be neglected.


A. How much work was done on the can by the launching device?
Express your answer in terms of some or all of the variables M , T , α0 and acceleration due to gravity g .

WA =

B. How much work is done on the can if it is launched at the same angle but stays in the air twice as long?
Express your answer in terms of some or all of the variables M , T , α0 and acceleration due to gravity g .
WB =

C. How does your result in part B compare to the answer to part A?
Express your answer to two significant figures.
WB/WA =

Answer 1

Hi there!

A.

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
$T_h = \frac{T}{2}$

Now, we can determine the velocity at which the can was launched at using the following equation:
$v_f = v_i + at$

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
$0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}$

***vsinθ is the vertical component of the velocity.

Solve for 'v':
$vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}$

Now, recall that:
$W = \Delta KE = \frac{1}{2}m(\Delta v)^2$

Plug in the expression for velocity:
$W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}$

B.

We can use the same process as above, where T' = 2T and Th = T.

$v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}$

C.

The work done in part B is 4 times greater than the work done in part A.

$\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}$