Answer to Question #153204 in Physics for Charles

Question

A 5kg block was placed on an inclined plane(50 degrees with respect to the vertical).Forces A is at angle C with respect to the horizontal. Force B is parallel with the incline.The coefficients of static and kenetic friction are 0.3 and 0.2 respectively.

A=197KN

B=49 KN

C=7 degrees

A.Determine if there is motion and its direction

The block is initially 5m from the ground (vertically).Based on your answer in "A",if block will be moved 2m vertically from its original elevation (relative to the incline and depending on the direction of motion)in a span of 5 seconds.

How much work was done?

Answer 1

A.

$N=mg\cos40°+197000\cdot\sin47°=5\cdot9.8\cdot\cos40°+197000\cdot\sin47°=$

$=144114(N)$

$49000+mg\sin40°-197000\cos47°=49000+5\cdot9.8\cdot\sin40°-197000\cos47°=$

$=-85322(N)$

$F_f=0.3N=0.3\cdot144114=43234(N)$

$F_{total}=-85322+43234=-42088(N)$

The block will move up the inclined plane. Answer

B.

$W=Fs=42088\cdot\frac{2}{\sin40°}=131(kJ)$ . Answer