What are those that you are wearing?
1 answer:
You might be interested in
Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
Answer:
:Final block velocity (toward the left)
:Final mass velocity (to the right)
Explanation:
To solve this problem we apply the theory of shocks:
In an elastic shock the kinetic energy and the amount of linear movement or momentum are conserved.
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.
Principle of conservation of the momentum:
m1vi1+m2vi2=m1vf1+m2vf2 Equation 1
Formula to calculate the coefficient of elastic restitution (e):
Equation 2
m1: Block mass
m2: mass of the body that collides with the block
Vi1,vf1: initial, final velocity of the block
Vi2,vf2: initial, final velocity of the body that collides with the block
Of the problem data we know that:
m1=m , m2 = 2m, vi1=v and vi2=0, then, we replace this information in equation (1) :
mv+0=mvf1+2mvf2 we divide by m:
v=vf1+2vf2 Equation (3)
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1,then , we we replace this information in equation (2)
v=vf2-vf1 Equation (4)
vf2=v+vf1 Equation (5)
We replace the equation 5 in the equation (3)
v=vf1+2(v+vf1)
v=vf1+2v+2vf1
-3vf1=v
We replace Vf1=(-1/3)v in the equation (5):
Answer;
:Final block velocity (toward the left)
:Final mass velocity (to the right)