Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.

Now here we will do the components of the weight of the person

given that weight of the person = 500 N

now its components are

$W_x = 500 cos37$

$W_y = 500 sin37$

now here as we can say that one of the component is balanced here by the normal force perpendicular to plane

while the other component of the weight is balanced by the force applied on the rope

So here the force applied on the rope will be given as

$F = W_y = 500 sin37$

$F = 300 N$

so it apply 300 N force along the inclined plane

5 0

3 years ago

1. The speed limit on a particular freeway is 28.0 m/s (about 101 km/hour). A car

zavuch27 [327]

Answer:

the speed limit 6 seconds and the car will travel in 90m

3 0

3 years ago

A heat engine with a thermal efficiency of 45 percent rejects 500 kj/kg of heat. how much heat does it receive

adell [148]

Alot as far as i know unless you need it in formal terms.

8 0

3 years ago