A 2.44 kg block is pushed 1.55 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle o
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a)
b) 5.2 m/s
c) 151.2 N
Explanation:
a)
When the box is on the frictionless ramp, there is only one force acting in the direction along the ramp: the component of the forc of gravity parallel to the ramp, which is given by
where
m =16 kg is the mass of the box
is the acceleration due to gravity
is the angle of the ramp
According to Newton's second law of motion, the net force on the box is equal to the product of mass and acceleration, so:
where a is the acceleration.
From the equation above we get
And we are told that the acceleration must not exceed
Substituting this value and solving for , we find the maximum angle of the ramp:
b)
Here we are told that the vertical distance of the ramp is
Since there are no frictional forces acting on the box, the total mechanical energy of the box is conserved: this means that the initial gravitational potential energy of the box at the top must be equal to the kinetic energy of the box at the bottom of the ramp.
So we have:
where:
m = 16 kg is the mass of the box
h = 1.4 m height of the ramp
v = final speed of the box at the bottom of the ramp
Solving for v,
c)
There are two forces acting on the box in the direction perpendicular to the ramp:
- The normal force, N, upward
- The component of the weight perpendicular to the ramp, downward, of magnitude
Since the box is in equilibrium along the perpendicular direction, the net force is zero, so we can write:
and by substituting:
m = 16 kg
We can find the normal force: