A 7.5kg block is sliding down a wall with constant velocity. The coefficient of static friction between the block and the wall
1 answer:
Ok, a couple of things have to be accounted for here. First, since the block is moving relative to the wall we have to use the kinetic coefficient of friction, 0.40. The second consideration is that since the block is moving at a constant velocity, the acceleration is zero. This means, by Newton's second Law, that the net force is zero. So the force of gravity must be equal to the friction force of the wall resisting its fall. This friction force is the product of the normal force (which we are seeking) and the kinetic coefficient of friction. We can then set these two forces equal:
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Answer:
q₃=5.3nC
Explanation:
First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:
Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:
In words, the value of q₃ must be 5.3nC.
Complete question:
A child sits in a wagon with a pile of 0.64-kg rocks. If she can throw each rock with a speed of 7.5 m/s relative to the ground, causing the wagon to move, how many rocks must she throw per minute to maintain a constant average speed against a 3.9-N force of friction
Answer:
The number of rocks per minutes thrown is 49 rocks/min
Explanation:
Given;
mass of the rock, m = 0.64 kg
speed of the rock, v = 7.5 m/s
frictional force, = 3.9 N
For an object to move at a constant speed, the applied force must be equal to the frictional force.
where;
is the applied force
N/t is the number of rocks per minutes thrown
Substitute the given parameters;
Approximately 49 rocks/min