Question

A 3.00-kg block starts from rest at the top of a 33.0° incline and slides 2.00 m down the incline in 1.80 s. (a) Find the acceleration of the block. 1.23 Correct: Your answer is correct. m/s2 (b) Find the coefficient of kinetic friction between the block and the incline. .23 Incorrect: Your answer is incorrect. (c) Find the frictional force acting on the block. N (d) Find the speed of the block after it has slid 2.00 m.

Answer 1

(a) $1.23 m/s^2$

Let's analyze the motion along the direction of the incline. We have:

- distance covered: d = 2.00 m

- time taken: t = 1.80 s

- initial velocity: u = 0

- acceleration: a

We can use the following SUVAT equation:

$d = ut + \frac{1}{2}at^2$

Since u=0 (the block starts from rest), it becomes

$d=\frac{1}{2}at^2$

So by solving the equation for a, we find the acceleration:

$a=\frac{2d}{t^2}=\frac{2(2.00 m)}{(1.80 s)^2}=1.23 m/s^2$

(b) 0.50

There are two forces acting on the block along the direction of the incline:

- The component of the weight parallel to the surface of the incline:

$W_p = mg sin \theta$

where

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration due to gravity

$\theta=33.0^{\circ}$ is the angle of the incline

This force is directed down along the slope

- The frictional force, given by

$F_f = - \mu mg cos \theta$

where

$\mu$ is the coefficient of kinetic friction

According to Newton's second law, the resultant of the forces is equal to the product between mass and acceleration:

$W-F_f = ma\\mg sin \theta - \mu mg cos \theta = ma$

Solving for $\mu$, we find

$\mu = \frac{g sin \theta - a}{g cos \theta}=\frac{(9.8 m/s^2)sin 33.0^{\circ} - 1.23 m/s^2}{(9.8 m/s^2) cos 33.0^{\circ}}=0.50$

(c) 12.3 N

The frictional force acting on the block is given by

$F_f = \mu mg cos \theta$

where

$\mu = 0.50$ is the coefficient of kinetic friction

m = 3.00 kg is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

$\theta=33.0^{\circ}$ is the angle of the incline

Substituting, we find

$F_f = (0.50)(3.00 kg)(9.8 m/s^2) cos 33.0^{\circ} =12.3 N$

(d) 6.26 m/s

The motion along the surface of the incline is an accelerated motion, so we can use the following SUVAT equation

$v^2 - u^2 = 2ad$

where

v is the final speed of the block

u = 0 is the initial speed

a = 1.23 m/s^2 is the acceleration

d = 2.00 m is the distance covered

Solving the equation for v, we find the speed of the block after 2.00 m:

$v=\sqrt{u^2 + 2ad}=\sqrt{0^2+2(9.8 m/s^2)(2.00 m)}=6.26 m/s$