A 42-kg box slides down a frictionless inclined plane. The angle of the inclined plane is 34.0 degrees. The distance from the box to the bottom of the inclined plane is 3 m. Draw the free body diagram of the force vectors acting on the box. Determine:
a. The weight of the box
b. the normal force acting on the box
c. the acceleration of the box down the incline
d. the velocity of the box a the bottom of the inclined plane
e. the time it took for the box to slide to the bottom of the inclined plane
a 42*9.81
b 42*9.81 cos 34
c force down slope = 42*9.81 * sin 34
so
a = 9.81 sin 34
d 3 = .5 (9.81*sin34) * t^2
t is answer to e
v = a t = 9.81 sin 34 *t
A box weighing 20lb. is being pushed up an incline. If the angle of incline was 90° to the ground, what effort would be needed to push the mass?
To determine the answers to the given questions, let's start by drawing the free body diagram of the force vectors acting on the box:
1. Weight of the box (a):
The weight of an object can be calculated using the formula: weight = mass * gravity, where gravity is approximately 9.8 m/s² on Earth. Given that the mass of the box is 42 kg, we can calculate its weight as follows:
Weight = mass * gravity = 42 kg * 9.8 m/s² = 411.6 N.
2. Normal force acting on the box (b):
The normal force is the force exerted by a surface to support the weight of an object resting on it. Since the inclined plane is free of friction, the normal force will be perpendicular to the plane and equal in magnitude but opposite in direction to the weight of the box. Therefore, the normal force will also be 411.6 N.
3. Acceleration of the box down the incline (c):
The acceleration of the box down the incline can be calculated using the formula: acceleration = gravitational force * sine(theta), where theta is the angle of the inclined plane. So,
Acceleration = weight * sin(theta) = 411.6 N * sin(34.0°) = 233.6 N.
4. Velocity of the box at the bottom of the inclined plane (d):
To find the final velocity of the box at the bottom of the inclined plane, we can use the SUVAT equation: v² = u² + 2as, where v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration (which we calculated as 233.6 N), and s is the distance (3 m). Rearranging the equation, we have:
v² = 0 + 2 * 233.6 m/s² * 3 m = 1401.6 m²/s².
Taking the square root of both sides, we find the velocity of the box at the bottom of the incline is:
v = sqrt(1401.6) m/s ≈ 37.4 m/s.
5. Time it took for the box to slide to the bottom of the inclined plane (e):
Time can be calculated using the equation: time = distance / velocity. Plugging in the values, we have:
Time = 3 m / 37.4 m/s ≈ 0.0802 s.
Therefore, the answers are:
a. The weight of the box is 411.6 N.
b. The normal force acting on the box is 411.6 N.
c. The acceleration of the box down the incline is 233.6 m/s².
d. The velocity of the box at the bottom of the incline is approximately 37.4 m/s.
e. The time it took for the box to slide to the bottom of the incline is approximately 0.0802 seconds.