5. In a local bar, a customer slides an empty beer mug on the counter for a refill. The bartender does not see the mug, which sl
1 answer:
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As we know that block of steel is continue to be in moving state
so here the friction must be kinetic friction between two surface
so we know that formula of kinetic friction must be
now we have
now from the above equation we have
So here we need atleast 14.25 N force to continue sliding the box now as per given options all forces which are less than 14.25 N is not correct
Hence correct answer must be 18.0 N
<u>D) 18.0 N</u>
The correct options are: B, C and D.
In fact:
- B is true, because the motion of the projectile can be decomposed in two independent motions on the x and y axis. On the x-axis, there is no acceleration; while on the y-axis, the projectile is accelerating towards the ground with
(acceleration of gravity)
- C is true, because the two motions on the horizontal and vertical direction are independent. In particular, the horizontal motion is a uniform motion (constant velocity), while the vertical motion is a uniformly accelerated motion (because of the gravitational acceleration g acting on the projectile)
- D is true, in fact (if we neglect air resistance) gravity is the only force acting on the projectile.
In fact:
- B is true, because the motion of the projectile can be decomposed in two independent motions on the x and y axis. On the x-axis, there is no acceleration; while on the y-axis, the projectile is accelerating towards the ground with
- C is true, because the two motions on the horizontal and vertical direction are independent. In particular, the horizontal motion is a uniform motion (constant velocity), while the vertical motion is a uniformly accelerated motion (because of the gravitational acceleration g acting on the projectile)
- D is true, in fact (if we neglect air resistance) gravity is the only force acting on the projectile.
Answer:
b. mg ( μ · cos θ + sin θ)
Explanation:
Hi there!
Please, see the attached figure for a graphical description of the problem.
We have the following parallel forces acting on the block (in parallel direction to the direction of movement):
F = applied force.
Fr = friction force.
wx = parallel component of the weight.
According to Newton´s second law:
∑F = m · a
Where "m" is the mass of the block and "a" its acceleration.
Then:
F - Fr - wx = m · a
Since the block is to be pushed at a constant velocity, the acceleration is zero. Then:
F - Fr - wx = 0
F = Fr + wx
The applied force has to be equal to the friction force plus the parallel component of the weight to push the block at constant velocity.
The friction force is calculated as follows:
Fr = μ · N
Where N is the normal force and μ is the coefficient of friction.
Notice that the normal force is of the same magnitude as the perpendicular component of the weight, wy.
Let´s apply Newton´s second law in the perpendicular direction to show this:
∑F = m · a
N - wy = m · a
The acceleration of the block in the perpendicular direction is zero. Then:
N - wy = 0
N = wy
And wy can be obtained by trigonometry (see figure):
wy = W · cos θ
N = wy = mg · cos θ
The parallel component of the weight is calculated using trigonometry (see figure):
wx = W · sin θ
wx = mg · sin θ
Then the applied force will be:
F = Fr + wx
F = μ · N + mg · sin θ (N = wy = mg · cos θ)
F = μ · mg · cos θ + mg · sin θ
F = mg ( μ · cos θ + sin θ)
The correct answer is the b.
