# Force of friction work problem

1. Mar 28, 2005

### the_d

A 19.1 kg block is dragged over a rough, hor-
izontal surface by a constant force of 195 N
acting at an angle of angle 29.1 degrees above the
horizontal. The block is displaced 30.3 m,
and the coeficient of kinetic friction is 0.116.
The acceleration of gravity is 9.8 m/s^2 .

the question ask for he magnitude of work done by the force of friction.

I found the force of friction to be 21.71 N and but the answer i'm getting for Work of frictional force is wrong. i dont know whats the matter, am i reading the problem wrong or sumtin? Work of Frictional force = Ffr(x)(cos180) i used 30.3 for x like the problem gave.

2. Mar 28, 2005

### xanthym

{Work by Friction} = {Friction Force}*{Distance} =
= {Net Normal Force on Surface}*{Coeff of Friction}*{Distance}
= {(Block Weight) - (Normal Component Pulling Force)}*{Coeff of Friction}*{Distance} =
= {(19.1 kg)*(9.8 m/sec^2) - (195 N)*sin(29.1 deg)}*(0.116)*(30.3 m) =
= (324.6 J)

~~

Last edited: Mar 28, 2005
3. Mar 28, 2005

### plusaf

so?

if the block weighs 19.1 kg and the force is not parallel to the horizontal surface, what are all of the forces acting on the block? vertical components of gravity, the block's mass, and the "pulling force", and the horizontal components of the frictional force from the surface and the "pulling force", have to balance, right? if not, the block will be accelerating.

one way or the other, if w=f*d, the frictional force times the 30.3m should equal the total work done by the frictional force. do you have the "right answer" supplied to you?

4. Mar 28, 2005

### Staff: Mentor

You got this answer by assuming that the normal force equals the weight. Not true, since the applied force changes the normal force. Instead, solve for the normal force by applying the equilibrium condition:
$\Sigma F_y = 0$
$F_n - mg + F_{applied}sin\theta = 0$
Once you solve for $F_n$, the friction will be $\mu F_n$.