the rate of work done or doing work is usually measured by the power of a machine .
A) 
Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:
Parallel:
(1)
where
m is the mass
g = 9.8 m/s^2 the acceleration of gravity

is the coefficient of friction
R is the normal reaction
a is the acceleration
Perpendicular:
(2)
From (2) we find

And substituting into (1)

Solving for a,

B) 5.94 m/s
We can solve this part by using the suvat equation

where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
Here we have
v = ?
u = 0 (it starts from rest)

s = 8.70 m
Solving for v,

Answer:
https://young.scot/get-informed/national/gender-identity-terms
Explanation:
It’s A.Longitudinal. Tell me if I’m right