Q2. A block of massmrests on the left edge of a block of larger massM. The coefficient of kinetic
friction between the two blocks is �, and the surface on which the larger block rests is frictionless.
A constant horizontal force of magnitude F is applied to the block of mass m, setting it in motion
as shown in figure below.
(a) Draw a free body diagram indicating all the forces acting on mass m.
(b) What is the acceleration of the mass m?
(c) Draw a free body diagram indicating all the forces acting on mass M.
(d) What is the acceleration of mass M?
(e) How long will it take the block of mass m to cover a distance L on the larger block?
(f) How far does the larger block moves during the time calculated in part (a)?
2
(a) The free body diagram for mass m would include the following forces:
- Force due to gravity (mg) acting downwards
- Normal force (N) exerted by the larger block on mass m, acting upwards
- Kinetic friction force (Fk) opposing the motion, acting to the left
- Applied force (F) acting to the right
The free body diagram for mass m would look like this:
N
↑
|
<--- Fk F --->
|
↓
mg
(b) To find the acceleration of mass m, we need to use Newton's second law of motion. The net force acting on mass m is given by:
Fnet = F - Fk
Using the coefficient of kinetic friction, we can express the kinetic friction force as:
Fk = �N
The normal force can be calculated as the force exerted downwards by the larger block, which is equal to the weight of mass m:
N = mg
Substituting these values into the equation for net force, we get:
Fnet = F - �N = F - �mg
According to Newton's second law, Fnet = ma, where a is the acceleration of mass m. Therefore, we have:
ma = F - �mg
Solving for acceleration, we get:
a = (F - �mg) / m
(c) The free body diagram for mass M would include the following forces:
- Force due to gravity (Mg) acting downwards
- Normal force (N) exerted by the surface underneath, acting upwards
The free body diagram for mass M would look like this:
N
↑
|
↓
Mg
(d) Since the surface on which mass M rests is frictionless, there is no friction force acting on mass M. Therefore, the acceleration of mass M would be zero.
(e) To find the time it takes for mass m to cover a distance L on the larger block, we can use the equation of motion:
S = ut + (1/2)at^2
Here, S is the distance covered (L), u is the initial velocity (which is zero), a is the acceleration of mass m, and t is the time we want to find.
Simplifying the equation, we get:
L = (1/2)at^2
Solving for t, we get:
t = √(2L/a)
Substituting the expression for acceleration from part (b), we get:
t = √(2L / [(F - �mg) / m])
(f) Since the acceleration of mass M is zero (as concluded in part (d)), the larger block does not move during the time calculated in part (a). Therefore, the distance covered by the larger block would be zero.
(a) To draw the free body diagram for mass m, we need to consider all the forces acting on it.
The force applied to m is F. There is also the gravitational force acting downwards, which we'll denote as mg, where g is the acceleration due to gravity. Additionally, there is a frictional force opposing the motion of m, which we'll denote as f_k.
Therefore, the free body diagram for mass m would include the following forces:
- Applied force F (horizontally to the right)
- Gravitational force mg (vertically downwards)
- Frictional force f_k (horizontally to the left)
(b) To find the acceleration of mass m, we can use Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration (F_net = ma).
The net force acting on mass m is the force applied (F) minus the frictional force (f_k). So, F_net = F - f_k.
Using the equation F_net = ma, we can rearrange it to find the acceleration:
F - f_k = ma
Therefore, the acceleration of mass m, denoted as a_m, is given by:
a_m = (F - f_k) / m
(c) To draw the free body diagram for mass M, we need to consider all the forces acting on it.
Since the surface on which mass M rests is frictionless, there is no frictional force acting on M. The only force acting on M is the gravitational force exerted on it, which we'll denote as M * g, where g is the acceleration due to gravity.
Therefore, the free body diagram for mass M would include the following force(s):
- Gravitational force M * g (vertically downwards)
(d) Since there is no net external force acting on mass M in the horizontal direction (as there is no frictional force), the acceleration of mass M, denoted as a_M, is zero. This means that mass M remains at rest or moves with a constant velocity.
(e) To find the time it takes for mass m to cover a distance L on the larger block, we can use the equation of motion:
L = (1/2) * a_m * t^2
Rearranging the equation gives:
t = sqrt((2L) / a_m)
Substituting the previously calculated acceleration a_m will give you the time it takes for mass m to cover distance L on the larger block.
(f) Since mass M moves with zero acceleration, its displacement during the time calculated in part (e) would be zero. Therefore, the larger block does not move during this time period and its displacement is zero.