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Masja [62]
3 years ago
15

A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down

the incline? show all work please​
Physics
1 answer:
Leona [35]3 years ago
6 0

Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

wsin\theta-\mu F_n=ma That's everything we need.

w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

2.0 × 10¹ - 8.8 = 6.0a and

11 = 6.0a so

a = 1.8 m/s/s

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Answer:0.27

Explanation:

Given

One worker Pushes with force F_1=430 N

other Pulls it with a rope of rope F_2=360 N

mass of crate m=290 kg

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

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