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A body of mass 1 kg initially at rest slides down an inclined plane that is 1 m high and 10 m long if the body experience a constant resistive force of 0.5 N while traveling on the slope what is the kinetic energy of the body at the base of the plane

User Garet
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Final answer:

The kinetic energy of the 1 kg body at the base of the 1 m high and 10 m long inclined plane, after overcoming a constant resistive force of 0.5 N, is 4.8 J.

Step-by-step explanation:

To calculate the kinetic energy of the body at the base of the inclined plane, we need to consider the work-energy principle, which states that the work done by all the forces acting on an object is equal to the change in kinetic energy of the object. When the body slides down the incline, it is acted upon by gravity, which does positive work, and a resistive force, which does negative work.

If h is the height of the incline, the gravitational potential energy at the top is mgh (where g is the acceleration due to gravity, 9.8 m/s2). As the body slides down the plane, this potential energy is converted into kinetic energy and work done against the resistive force. If d is the length of the incline and f is the constant resistive force, then the work done against the resistive force is f × d.

Thus, if m is the mass of the body, the total work W done on the body is: W = mgh - f × d. The kinetic energy (KE) of the body at the base of the plane is equal to the work done, W. Substituting the given values, we have m = 1 kg, h = 1 m, d = 10 m, f = 0.5 N, and g = 9.8 m/s2. The kinetic energy at the base of the plane is thus KE = mgh - f × d = (1 kg)(9.8 m/s2)(1 m) - (0.5 N)(10 m) = 9.8 J - 5 J = 4.8 J.

User RaviRokkam
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