(a) [tex]1.23 m/s^2[/tex]
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:
[tex]d = ut + \frac{1}{2}at^2[/tex]
Since u=0 (the block starts from rest), it becomes
[tex]d=\frac{1}{2}at^2[/tex]
So by solving the equation for a, we find the acceleration:
[tex]a=\frac{2d}{t^2}=\frac{2(2.00 m)}{(1.80 s)^2}=1.23 m/s^2[/tex]
(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:
[tex]W_p = mg sin \theta[/tex]
where
m = 3.00 kg is the mass of the block
g = 9.8 m/s^2 is the acceleration due to gravity
[tex]\theta=33.0^{\circ}[/tex] is the angle of the incline
This force is directed down along the slope
- The frictional force, given by
[tex]F_f = - \mu mg cos \theta[/tex]
where
[tex]\mu[/tex] 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:
[tex]W-F_f = ma\\mg sin \theta - \mu mg cos \theta = ma[/tex]
Solving for [tex]\mu[/tex], we find
[tex]\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[/tex]
(c) 12.3 N
The frictional force acting on the block is given by
[tex]F_f = \mu mg cos \theta[/tex]
where
[tex]\mu = 0.50[/tex] 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
[tex]\theta=33.0^{\circ}[/tex] is the angle of the incline
Substituting, we find
[tex]F_f = (0.50)(3.00 kg)(9.8 m/s^2) cos 33.0^{\circ} =12.3 N[/tex]
(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
[tex]v^2 - u^2 = 2ad[/tex]
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:
[tex]v=\sqrt{u^2 + 2ad}=\sqrt{0^2+2(9.8 m/s^2)(2.00 m)}=6.26 m/s[/tex]
