A 100g block lies on an inclined plane that makes an angle of 15 degrees with the horizontal. The coefficient of kinetic frictio
n between the block and the plane is 0.6. Attached to this block is a massless, frictionless rope connected to a block-and-tackle type device (again, frictionless and massless). Hanging on the pulley is a container of mass 30g. How much mass should you put in the container so that the 100g block slides down the inclined plane at constant speed
1 answer:
Answer:
Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g
Explanation:
The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.
And sum of upward forces = sum of downward forces.
N = mg cos θ
m = 100 g = 0.10 kg
g = acceleration due to gravity = 9.8 m/s²
θ = 15°
N = (0.1×9.8×cos 15°) = 0.946582 N
The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).
For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.
Frictional force = Tension + F
Frictional force = μN
where μ = coefficient of kinetic friction = 0.60
N = normal reaction = 0.9466 N
Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N
The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ
F = (0.10 × 9.8 × sin 15°) = 0.253624 N
Tension in the rope = T = ?
Fr = F + T
T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N
But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.
2T = mg
2 × 0.3144 = 9.8m
m = 0.06416 kg = 64.16 g.
Mass of the container = 30 g
So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g
Hope this Helps!!!
You might be interested in
Answer:
75 m
Explanation:
The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.
The horizontal component of the velocity of the projectile is
and it is constant during the motion;
the total time of flight is
t = 5 s
Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:
Complete Image
Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by and
, respectively.The ratio ΔLthick rod/ΔLthin rod is:
a.) =1
b.) <1
c.) >1
Answer:
The ratio is less than 1 i.e option B is correct
Explanation:
The Young Modulus of a material is generally calculated with this formula
Where is the stress =
is the strain =
Making Strain the subject
now in this question we are that the same tension was applied to both wires so
would be constant
Hence
for the two wire we have that
Looking at young modulus formula
Now we are told that a comprehensive force is applied to the wire so for this question
is constant
And given that the length are the same
so
Now we are told that one is that one rod is twice as thick as the other
So it implies that one would have an area that would be two times of the other
Assuming that
So
From the question the length are equal
So
Hence the ratio is less than 1