Verified Answer :

Rochirion

Answer:

C. 42.26N

Explanation:

To find the frictional force keeping a block at rest on a 25° inclined plane, we need to resolve the forces acting on the block and identify the component of the gravitational force parallel to the incline.

We can sum the forces acting in the x-prime direction to determine the static frictional force:

\(\sum F_{x'}: \vec f_s+\vec w \cos (270^\circ-25^\circ)=0\)

\(\Longrightarrow \vec f_s=-\vec w \cos (245^\circ)\)

The weight of the block is given as 100 N:

\(\Longrightarrow \vec f_s=-(100 \text{ N})\cos (245^\circ)\\\\\\\\\therefore \vec f _ s \approx \boxed{42.26 \text{ N}}\)

Thus, the correct option is C.

View image Rochirion

The correct option is c. 42.26N. The frictional force keeping the block at rest on a 25° inclined plane with a weight of 100 N is approximately 42.26 N.

To calculate the frictional force keeping the block at rest on an inclined plane, we need to resolve the forces acting on the block. The weight of the block (W) is 100 N, and the ramp is inclined at 25°. The components of the weight are:

Wparallel = W sin(25°)

Wperpendicular = W cos(25°)

Given:

  • W = 100 N
  • θ = 25°

Calculating the components:

Wparallel = 100 N sin(25°) ≈ 42.26 N

Wperpendicular = 100 N cos(25°) ≈ 90.63 N

The block is at rest, implying that the static frictional force (fs) equals the parallel component of the weight:

fs = Wparallel ≈ 42.26 N

Therefore, the frictional force keeping the block at rest is 42.26 N.

Related Questions