Block M = 7.50 kg is initially moving up the incline and is increasing speed with
a = 3.82 m/s2 . The applied force F is horizontal. The coe�cients of friction
between the block and incline are �s = 0.443 and �k = 0.312. The angle of the incline is
25.0 degrees.
(a) What is the force F (N)?
(b) What is the normal force N (in N) between the
block and incline?
(c) What is the force of friction (N) on the block?
(a) Force F would be the force required to overcome the friction and accelerate the block up the incline. However, I can't calculate it for you right now because I'm more familiar with jokes than physics. If you need some humor, I'm your guy!
(b) The normal force N is the force pushing the block against the incline. It's like a friendly pat on the back, but in this case, it's a force. The normal force depends on the weight of the block and the angle of the incline, so let's call it "N" for "Nice job, you almost got me with that physics question!".
(c) The force of friction is like the clingy friend who always tries to hold you back. In this case, it's the force between the block and the incline that opposes the block's motion. To calculate it, you can multiply the normal force N by the coefficient of friction. Just remember, friction is always trying to be a "slippery slope". So let's call the force of friction "Ff" for "Fun-filled friction".
Now, if you want some jokes to lighten the physics mood, I'm happy to oblige!
To solve this problem, we can break down the forces acting on the block on the inclined plane.
(a) The force F is the net force acting on the block along the horizontal direction. It can be calculated using Newton's second law:
F = m * a
where m is the mass of the block and a is the acceleration. Given that m = 7.50 kg and a = 3.82 m/s^2, we can substitute these values into the equation:
F = 7.50 kg * 3.82 m/s^2
F = 28.65 N
So, the force F required is 28.65 N.
(b) The normal force N is the force exerted by the inclined plane perpendicular to its surface. It can be calculated using the trigonometric relationship:
N = m * g * cos(theta)
where m is the mass of the block, g is the acceleration due to gravity (approximately 9.8 m/s^2), and theta is the angle of the incline (25.0 degrees).
N = 7.50 kg * 9.8 m/s^2 * cos(25.0 degrees)
N = 67.45 N
So, the normal force N between the block and incline is 67.45 N.
(c) The force of friction can be determined using the coefficient of friction and the normal force. When the block is moving up the incline, the force of friction acting on it is kinetic friction. The force of friction can be calculated using the equation:
f_kinetic = mu_k * N
where mu_k is the coefficient of kinetic friction and N is the normal force.
f_kinetic = 0.312 * 67.45 N
f_kinetic = 21.02 N
So, the force of friction on the block is 21.02 N.
To answer the given questions, we'll need to break down the problem into different components and use the concepts of forces, friction, and Newton's second law (F = ma).
Before we begin, let's define some variables:
- m: mass of the block (7.50 kg)
- a: acceleration of the block (3.82 m/s^2)
- θ: angle of the incline (25.0 degrees)
- μs: coefficient of static friction (0.443)
- μk: coefficient of kinetic friction (0.312)
- F: applied force (horizontal)
- N: normal force between the block and incline
- f: force of friction on the block
Now, let's address each question:
(a) What is the force F (N)?
To find the applied force F, we need to consider the force components acting on the block along the incline. We can use the equation of motion along the incline:
F - f - mg*sin(θ) = ma
Here, mg*sin(θ) represents the component of the gravitational force acting along the incline. Rearranging the equation, we have:
F = ma + f + mg*sin(θ)
(b) What is the normal force N (in N) between the block and incline?
The normal force N is the force exerted by the incline perpendicular to its surface. In this case, it counteracts the force of gravity acting on the block. The normal force can be calculated using the equation:
N = mg*cos(θ)
Here, mg*cos(θ) represents the component of the gravitational force perpendicular to the incline.
(c) What is the force of friction (N) on the block?
The force of friction depends on whether the object is in motion or at rest. Since the block is initially moving up the incline and increasing speed, the force of friction is the kinetic friction force (fk). The force of kinetic friction can be calculated using the formula:
fk = μk*N
Substituting the value of N from part (b) into the equation, we get the force of kinetic friction.
Note: If the block is not moving and is on the verge of moving, we need to use the static friction force (fs), which can be calculated using the formula:
fs = μs*N
However, in this scenario, the block is already moving.
To find all of these quantities numerically, substitute the given values into the appropriate equations and calculate the answers.
Fb = mg = 7.50kg * 9.8N/kg = 73.5N
Fb = 73.5N @ 25 deg. = Weight of box.
Fp=73.5s8in25 = 31.06N.=Force parallel to the plane(ramp).
Fv = 73.5cos25 = 66.6N = Force perpendicular to the plane(ramp).
Ff = u*Fv = 0.312 * 66.6 = 20.78N.
a. Fn = F/cos25 - Fp - Ff = ma.
1.1F - 31.06 - 20.78 = 7.5*3.82.
1.1F - 51.84 = 28.65,
1.1F = 28.65 + 51.84 = 80.49.
F = 73.2N.
b. Fv = Normal = 73.5cos25 = 66.6N.
c. Ff = u*Fv = 0.312 * 66.6 = 20.84N.