Question

A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of the incline, and its speed after it has traveled 7.40 mm to the bottom of the incline is 3.80 m/sm/s.What is the speed of the block when it is 3.40 m from the top of the incline?

Answer 1

The speed of the block when it is 3.40 m from the top of the incline is 7.24 m/s.

Step-by-step explanation:

To determine the speed of the block when it is 3.40 m from the top of the incline, we can use the concepts of constant acceleration and the equations of motion. Since the block is released from rest at the top of the incline, its initial speed is 0 m/s. Using the equation v^2 = u^2 + 2as, where v is the final speed, u is the initial speed, a is the acceleration, and s is the displacement, we can solve for the final speed. Plugging in the given values, we have:

v^2 = 0^2 + 2 * a * 3.4

3.8^2 = 2 * a * 3.4

14.44 = 6.8a

a = 2.13 m/s^2

Now, we can use the equation v = u + at to find the speed. Plugging in the values, we have:

v = 0 + 2.13 * 3.4

v = 7.24 m/s

Answer 2

2.58 m/s

Step-by-step explanation:

We can use the following equation of motion to find out the constant acceleration of the block:

`v^2 - v_0^2 = 2a\Delta s`

where v = 3.8 m/s is the final velocity after it has traveled 7.4 m,
`v_0 = 0` is the initial velocity of the block when it starts from rest, , and
`\Delta s = 7.4` is the distance traveled.

`3.8^2 - 0 = 2*a*7.4`

`14.44 = 14.8a`

`a = 14.44 / 14.8 = 0.98 m/s^2`

We can use the same motion equation to calculate block speed at the end of 3.4m track

`v_2^2 - 0 = 2*0.98*3.4`

`v_2^2 = 6.63`

`v_2 = √(6.63) = 2.58 m/s`