Question

A girl who has a mass of 45kg slides down a slide that is inclined at an angle of 45° with the horizontal if the coefficient of kinetic friction between the girl and the slide is 0.25 what is the magnitude of her acceleration?

Plz if Somone can write the steps thanks ​

Answer 1

Step-by-step explanation:

This is one-dimensional motion on an incline, which is always tons of fun :/

The equation you need for this is

w*sinθ - f = ma where w is the weight of the girl (NOT the mass), theta is the angle of inclination, f is the frictional force pulling back on the girl, m is the mass of the girl, and a is the acceleration she experiences under these conditions. We have the mass but we need the weight, and weight has its own equation:

`F_n=w=mg` where Fn is the normal force of the girl (which is the same as weight). Her normal force/weight is

w = 45(9.8) and we need 2 sig fig's for the answer so her weight is

w = 440N

f = μ
`F_n` which is the coefficient of kinetic friction times the normal force/weight of the girl:

f = (.25)(440) and we need 2 sig fig's for this as well so the frictional force is

f = 110

Putting it all together in the main equation looks like this:

440sin45 - 110 = 45a

Multiplication/division rules for sig fig's are different so we have to do them individually.

440sin45 will have 2 sig fig's in the answer:

440sin45 = 310; therefore:

310 - 110 = 45a. Again, because we have subtraction on the left and we will divide eventually, and the rules for both are different, we will subtract first, get that answer in the correct number of sig fig's and then divide that result by 45:

2.0 × 10² = 45a so

a = 4.4 m/s/s