Asolid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the bottom of a hill, starting from rest. if the angular speed of the disk is 4.27 rad/s at the bottom, how high did it start on the hill

The height attained by the disk on the hill is 3.57 m.

The given parameters;

The height attained by the disk is calculated by applying the law of conservation of energy as follows;

Thus, the height attained by the disk on the hill is 3.57 m.

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Height reached by the ball, h = 3.57 meters

Explanation:

It is given that,

Mass of the disk, m = 42 kg

Diameter of the disk, d = 3.2 m

Radius, r = 1.6 m

Angular speed of the disk,

The kinetic energy of the disk is equal to its potential energy. Using the conservation of energy as :

h = 3.57 meters

So, the solid disk will reach to a height of 3.57 meters.

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards