a body of massb20kg slides down a plane inclined at 30degree to the horizontal if a constant resting force of 40N act on the body it acceleration down the plane is what?
f=mgsin
ma=mg sin
20×a=20×10sin30
20a=100
a=50
See previous post.
using F=MA, Mg=Mgsin
in Relation we have;-
M=20kg, F=40N, a=?
:-(a) as subject formula,
:- a=f/m, 40/20=2
mgsin= 2*10sin, 20sin30degree
10.
To find the acceleration of the body sliding down the inclined plane, we can start by using Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).
In this case, we have the force acting on the body, which is the component of the gravitational force pulling it down the plane. This force can be calculated using the formula F = mg sinθ, where m is the mass of the body, g is the acceleration due to gravity (approximately 9.8 m/s^2), and θ is the angle of inclination of the plane.
Plugging in the given values, we have:
F = (20 kg) × (9.8 m/s^2) × sin(30°)
F = (20 kg) × (9.8 m/s^2) × (0.5)
F = 98 N
We also have the constant resisting force acting on the body, which is given as 40 N.
Now, we can equate the net force acting on the body to the mass of the body multiplied by its acceleration:
Net force = 98 N - 40 N
Net force = 58 N
F = ma
58 N = (20 kg) × a
Solving for a:
a = (58 N) / (20 kg)
a ≈ 2.9 m/s^2
Therefore, the acceleration of the body sliding down the plane is approximately 2.9 m/s^2.