Question

a block of mass 1kg is projected from the lowest point of along the inclined plane if g is equal to 10 M per second squared the retardation experienced by the block is

Answer 1

Please find the answer below:

Keeping in view that; a block of mass 1kg is projected from the lowest point of along the inclined plane if g is equal to 10 M per second squared the retardation experienced by the block is 15/√2.

Explanation:

Forces of retardation:

(i) mgsin(x) downward along the inclined plane.

(ii) Force of fiction;

ο Coefficient of static friction= u

ο Normal force; R=mgcos(x)

ο Angle of inclination=x=45°

Hence,

F=ma=mgsin(x)+uR

1(a)=mgsin(x)+umgcos(x)

a=(1)(10)(sin40°)+(0.5)(1)(10)(cos45°)

a=10/√2+5/√2

a=15/√2

Answer 2

Explanation:

Retardation to the block is provided by the following forces:

- mgsin(x) downward along the inclined plane

- Force of friction (also downward along the inclined plane, since it is opposing block's upward motion)

     u: Coefficient of static friction

     Normal force, R = mgcos(x)

     x = angle of inclination = 45o

Thus, F = ma = mgsin(x) + uR = mgsin(x) + umgcos(x)

                 (1)a = (1)(10)(sin45o) + (0.5)(1)(10)(cos45o)

                 a = 10/sqrt(2) + 5/sqrt(2) = 15/sqrt(2)

Hope this helps! :)