Question

a block is pushed up a frictionless 40 incline. if the initial velocity after the push is 5.00 m/s. how far along the incline does the box move before turning around and sliding back down? how much time does it take the box to return to its original position.

Answer 1

Answer:

d=1.982m, t=1.019s

Explanation:

There are different approaches we can take to solve this problem. You could either solve this by using conservation of energy or by taking a kinematic approach. I'll solve this by using kinematics. So, the very first thing we need to do in order to solve this is do a drawing of the situation so we can analyze it better. (See attached picture).

So, since we are talking about an inclined plane, we can see that the force of gravity is being split into an x and y components if we incline the axis of coordinates. Taking this into account we can see that:

$\sum F_{x}=ma_{x}$

Since there is no friction in our system, then the only force acting upon the box is the force of gravity, or weight. Since we are taking the upwards direction as the positive direction of movement, we can say that the force of gravity is excerting a negative influence on our box, so this acceleration will be negative, so our sum of forces will now look like this:

$-mg sin(40^{o})=ma$

we can cancel the masses out so we can see that:

a=-g sin(40°)

$a=-9.81m/s^{2} sin(40^{o})$

We have now enough information to solve our problem.

we can take the following equation to find the distance the block travels up the incline:

$x=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

we know the final velocity must be zero, so we can use the provided data to solve our formula:

$x=\frac{(0)^{2}-(5m/s)^{2}}{2(-9.81m/s^{2})sin 40^{o}}$

which yields:

x=1.982m

In order to find the time it takes for the block to return to its original position we can use the following formula:

$x=V_{0}t+\frac{1}{2}at^{2}$

since x=0 is the starting point we can use that to solve our equation:

$0=5t+\frac{1}{2}(-9.81sin 40^{o})t^{2}$

which simplifies to:

$0=5t-4.905t^{2}$

which can now be solved for t

t(5-4.905t)=0

t=0                and          5-4.905t=0

t=0                 and         $t=\frac{5}{4.905}=1.019$

so the time it takes the block to return to its original position is

t= 1.019s