An object of mass 3kg is held at rest on a rough plane. The plane is inclined at 30º to the horizontal and has a coefficient of friction of 0.2. The object is released, what acceleration does the object move with?

We need to use Newtons law F=ma going down the slope.
We can see that the only forces acting in this direction are the component of the weight and friction, so we have that: F = Wsin30 - μR = 3a
We have that μ = 0.2, W = 3g, R = Wcos30 = 3gcos30
Hence, F = 3gsin30 - 0.2(3gcos30) = 3a
and so, (1/2)g - (√3/10)g = a
Therefore, acceleration = (1/2 - √3/10)g

AD
Answered by Asha D. Maths tutor

3372 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

If y=3x^3e^x; find dy/dx?


Find the tangent to the curve y = x^2 + 3x + 2 that passes through the point (-1,0), sketch the curve and the tangent.


A curve C has the equation x^3 + 2xy- x - y^3 -20 = 0. Find dy/dx in terms of x and y.


The point P (4, –1) lies on the curve C with equation y = f( x ), x > 0, and f '(x) =x/2 - 6/√x + 3. Find the equation of the tangent to C at the point P , giving your answer in the form y = mx + c. Find f(x)


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning