A small block slides down an incline with a constant acceleration. The block is released from rest at the top of the incline. After it travels 5 m to the bottom its speed is 5 m/s. What is the speed, in m/s, of the block when it had traveled only 2.1 m from the top

v = 3.24 m/s

Explanation:

Since we don't have time, we can use the formula;

(Final distance - initial distance)/time = (initial velocity + final velocity)/2

Thus;

(x_f - x_i)/t = ½(v_xi + v_xf)

We are given;

x_i = 0 m

x_f = 5 m

v_xi = 0 m/s

v_xf = 5 m/s

Thus, plugging in the relevant values;

(5 - 0)/t = (0 + 5)/2

5/t = 5/2

t = 2 s

Using Newton's first law of motion, we can find the acceleration.

v = u + at

Applying to this question;

5 = 0 + a(2)

5 = 2a

a = 5/2

a = 2.5 m/s²

To get the speed, in m/s, of the block when it had traveled only 2.1 m from the top, we will use the formula;

v² = u² + 2as

v² = 0² + 2(2.5 × 2.1)

v² = 10.5

v = √10.5

v = 3.24 m/s

force tends to be low and long diustance masses are high