Question

A 23 kg block slides down a frictionless slope which is at angle θ = 29◦ . starting from rest, the time to slide down is t = 1.81 s. the acceleration of gravity is 9.8 m/s 2 .what total distance s did the block slide? answer in units of m.

Answer 1

The block slides a total distance of approximately 7.76 meters down the frictionless slope.

Step-by-step explanation:

To calculate the total distance s the 23 kg block slid, we can use the kinematic equation for motion under uniform acceleration, which is given by:

s = ut + ½at²

where u is the initial velocity, a is the acceleration, and t is the time.

The block starts from rest, so the initial velocity u is 0. The acceleration a down the slope can be found using the component of gravitational force along the slope, which is:

a = g · sin(θ)

So,

a = 9.8 m/s² · sin(29°) = 9.8 · 0.4848 ≈ 4.75 m/s²

Now we can plug in the values:

s = 0 · 1.81 s + ½ · 4.75 m/s² · (1.81 s)²

s ≈ ½ · 4.75 m/s² · 3.2761 s²

s ≈ 7.7602 m

Therefore, the block slides a total distance of approximately 7.76 meters.