Question

A block of 2kg is thrown up an inclined plane of inclination 45^.... ​

Answer 1

Answer:

${a} \: is \: the \: right \: answer$

Explanation:

$more \: to \: know \:$

Given,

m=2kg

θ=30

0

μ=0.5

g=10m/s

2

The contact force is F

c

is the resultant of frictional force and normal force.

From the free body diagram,

Normal force, N=mgcosθ

N=2×10×cos30

0

N=17.320N

The friction force, F

r

=μmgcosθ

F

r

=0.5×2××10×cos30

0

F

r

=8.660N

N and F

r

are perpendicular to each other.

The contact force is, F

c

=

N

2

+F

r

2

F

c

=

(17.320)

2

+(8.699)

2

F

c

=19.381N

We can say approximately 20 N