Question

In the figure, a block weighing 22.0 N is held at rest against a vertical wall by a horizontal force F→ of magnitude 60 N. The coefficient of static friction between the wall and the block is 0.55, and the coefficient of kinetic friction between them is 0.38. In six experiments, a second force P→ is applied to the block and directed parallel to the wall with these magnitudes and directions: (a)33 N, up, (b)11 N, up, (c)48 N, up, (d)62 N, up, (e)9.2 N, down, and (f)18 N, down. In each experiment, what is the frictional force on the block, including sign? Take the direction up the wall as positive, and down the wall as negative. Next, calculate the acceleration, including sign, of the block in each case. Note that acceleration is zero if the block does not move.
(g) What is the acceleration in (a)?
(h) What is the acceleration in (b)?
(i) What is the acceleration in (c)?
(j) What is the acceleration in (d)?
(k) What is the acceleration in (e)?
(l) What is the acceleration in (f)?

Answer 1

a)The static frictional force is, (0.55) (60) 33N The kinetic friction force is, =(0.38) (60) =22.8N For P-33N up The vertical resultant force is, Fu=P W resl tant = 33- 22 11N(up)< f Frictional force is greater than applied force, so, it will not move upwards or down wards so, the acceleration is zero. a 0 m/s2 b)For P-11N, up F W P resul tant = 22-11 =11N(down) f. Frictional force is greater than applied force, so, it will not move upwards or down wards so, the acceleration is zero. a 0 m/s c)For p-48 N up F =P-W resultant =48-22 26N (up) f Frictional force is greater than applied force, so, it will not move upwards or down wards so, the acceleration is zero. a 0 m/s d)For p 62N, up The vertical resultant force is., FntPW el tant =62-22 =40N (up)> f _ The net force is, resid tant net = 40-22.8 = 17.2N -

The acceleration is net a т (17.2) 22 9.8 = 7.66m/s2 e) For p-9.2N, down The vertical resultant force is., resultant =9.2+ 22 = 31.2N (down)> f. - The net force is, resi tan ner =31.2-22.8 = 8.4N The acceleration is net a = т (8.4 22 9.8 =3.74m/s2 f) For P 18Ndown The vertical resultant force is., resl ant -18+22 = 40N(down)> f The net force is, -fi resid tant net = 40-22.8 = 17.2N

The acceleration is ner a (17.2) 22 9.8 = 7.66m/s2