A 3.00 kg block starts from rest at the top of a 36.0'> incline and slides 2.00 m down the incline in 1.60 s.
(a) Find the acceleration of the block.
m/s2
(b) Find the coefficient of kinetic friction between the block and the incline.
(c) Find the frictional force acting on the block.
N
(d) Find the speed of the block after it has slid 2.00 m.
m/s
To solve this problem, we can use the following equations:
(a) To calculate the acceleration of the block, we can use the kinematic equation:
v = u + at,
where v is the final velocity, u is the initial velocity (which is 0 in this case since the block starts from rest), a is the acceleration, and t is the time. Here, we know that the block slides down the incline for 2.00 m in 1.60 s.
Let's solve for a:
2.00 = 0 + a(1.60)
2.00 = 1.60a
a = 2.00 / 1.60
a = 1.25 m/s^2
Therefore, the acceleration of the block is 1.25 m/s^2.
(b) To find the coefficient of kinetic friction between the block and the incline, we can use the equation:
frictional force = coefficient of friction * normal force.
The normal force can be calculated using the following equation:
normal force = mg * cos(theta),
where m is the mass of the block (3.00 kg) and theta is the angle of the incline (36.0').
Let's calculate the normal force:
normal force = 3.00 kg * 9.8 m/s^2 * cos(36.0')
Next, we need to calculate the frictional force, which is given by the equation:
frictional force = m * a,
where m is the mass of the block (3.00 kg) and a is the acceleration (1.25 m/s^2).
Let's set up and solve the equation:
m * a = coefficient of friction * normal force
3.00 * 1.25 = coefficient of friction * (3.00 * 9.8 * cos(36.0'))
Now, we can solve for the coefficient of friction.
Coefficient of friction = (3.00 * 1.25) / (3.00 * 9.8 * cos(36.0'))
(c) To find the frictional force acting on the block, we can use the equation:
frictional force = coefficient of friction * normal force.
Substituting the values for the coefficient of friction and the normal force we calculated earlier, we can now solve for the frictional force.
frictional force = coefficient of friction * (3.00 * 9.8 * cos(36.0'))
(d) Finally, to find the speed of the block after it has slid 2.00 m, we can use the equation:
v^2 = u^2 + 2as,
where v is the final velocity, u is the initial velocity, a is the acceleration (1.25 m/s^2), and s is the distance traveled (2.00 m).
Let's solve for v:
v^2 = 0 + 2(1.25)(2.00)
v^2 = 5
v = √5
Therefore, the speed of the block after it has slid 2.00 m is √5 m/s.
To answer each of these questions, we will use various equations of motion and the principles of friction. Let's break down each question step by step:
(a) Find the acceleration of the block.
To find the acceleration, we need to use the equation of motion that relates distance, time, and acceleration:
𝑑 = 𝑣𝑖𝑛𝑖𝑡 + 0.5𝑎𝑡^2
The initial velocity (𝑣𝑖𝑛𝑖) is 0 m/s because the block starts from rest. The distance (𝑑) is given as 2.00 m, and the time (𝑡) is given as 1.60 s. Plugging the values into the equation, we have:
2.00 = 0(1.60) + 0.5𝑎(1.60)^2
Simplifying this equation, we can solve for acceleration:
2.00 = 0 + 0.8𝑎(2.56)
2.00 = 2.048𝑎
Dividing both sides by 2.048, we find:
𝑎 = 2.00 / 2.048
Therefore, the acceleration of the block is 0.977 m/s^2.
(b) Find the coefficient of kinetic friction between the block and the incline.
To find the coefficient of kinetic friction, we need to use the equation:
𝑓_𝑘 = 𝜇_𝑘 𝑁
where 𝑓_𝑘 is the kinetic frictional force, 𝜇_𝑘 is the coefficient of kinetic friction, and 𝑁 is the normal force. The normal force can be calculated as:
𝑁 = 𝑚𝑔𝑐𝑜𝑠(𝜃)
where 𝑚 is the mass of the block and 𝑔 is the acceleration due to gravity. 𝑐𝑜𝑠(𝜃) is the cosine of the angle of the incline, which is 36.0'.
The mass of the block is given as 3.00 kg, and the acceleration due to gravity is approximately 9.8 m/s^2.
Plugging in the values, we have:
𝑁 = 3.00 × 9.8 × cos(36.0')
Next, we need the frictional force 𝑓_𝑘. We can calculate this force using Newton's second law of motion:
𝑓_𝑘 = 𝑚𝑎
where 𝑚 is the mass of the block and 𝑎 is the acceleration from part (a).
Plugging in the values, we have:
𝑓_𝑘 = 3.00 × 0.977
Finally, we can equate the two expressions for 𝑓_𝑘:
𝜇_𝑘(3.00 × 9.8 × cos(36.0')) = 3.00 × 0.977
Rearranging the equation to solve for 𝜇_𝑘, we find:
𝜇_𝑘 = (3.00 × 0.977) / (3.00 × 9.8 × cos(36.0'))
Therefore, the coefficient of kinetic friction between the block and the incline is the calculated value of 𝜇_𝑘.
(c) Find the frictional force acting on the block.
We can use the equation 𝑓_𝑘 = 𝜇_𝑘 𝑁 from part (b) to find the frictional force 𝑓_𝑘. Plugging in the known values, we can calculate 𝑓_𝑘.
(d) Find the speed of the block after it has slid 2.00 m.
To find the speed, we can use the equation of motion that relates initial velocity, acceleration, and distance:
𝑣^2 = 𝑣𝑖𝑛𝑖^2 + 2𝑎𝑑
The initial velocity 𝑣𝑖𝑛𝑖 is 0 m/s, the acceleration 𝑎 is given (from part (a)), and the distance 𝑑 is given as 2.00 m. Plugging in the values, we have:
𝑣^2 = 0 + 2(0.977)(2.00)
Simplifying, we have:
𝑣^2 = 3.908
Taking the square root of both sides, we find:
𝑣 ≈ √3.908
Therefore, the speed of the block after it has slid 2.00 m is approximately the calculated value of √3.908 m/s.