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
To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and the geometric relationships that allow us to find the volume of the figures presented.
For the particular case of the Cube with equal sides its volume is determined by
In the case of perforated material we have that its volume is given according to the cylindrical geometry, that is to say
In this way the net volume would be
We need to find the mass, but we have the Weight and Gravity so from Newton's second Law
PART A) From the relation of density as a unit of mass and volume we have to
PART B) To find the weight of the cube then we apply the ratio of
Answer:
Newton's Second Law of Motion
Explanation:
According to Newton's second law of motion, the change in velocity of a body is directly proportional to the force applied on it. Velocity is a vector quantity. It measures the magnitude of the speed as well as its direction.
F = m a
where, F is the applied force, m is the mass and a is the acceleration.
It can also be expressed as:
where, p = mv ( momentum)