Question

A horizontal force of 99 N pushes a 11 kg block up a frictionless incline that makes an angle of 41° with the horizontal.

(a) What is the normal force that the incline exerts on the block?
                   =N

(b) What is the acceleration of the block? (Hint: assume up the block as the positive direction.)
                 =m/s²

Answer 1

Fsin41 41 deg Fcos41 masin41 mgmgco941 41 deg

Let's take the incline as the x-axis

Consider all forces in the y-direction

$\sum F_{y}=0$

$N-mgcos41-Fsin41=0$

$N=mgcos41+Fsin41$

$N=11kg*9.81m/s^{2}*cos41+99N*sin41$

$N=11*9.81*cos41+99*sin41$

(a) ANSWER : ${\color{Red} N=146.39N}$

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Consider the block

Net Force = Mass*acceleration

$Fcos41-mgsin41=ma$

$99N*cos41-11kg*9.81m/s^{2}sin41=11kg*a$

$99*cos41-11*9.81sin41=11*a$

$3.921=11*a$

$\frac{3.921}{11}=a$

(b) ANSWER: ${\color{Red} a=0.356m/s^{2}}$

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Answer 2

Draw the free body diagram of the block. Fcoso mgcos Fsin+mgcose/ 7.--- mg The normal reaction is expressed from the figure as follows: N = mg cosO+F sino

Substitute 9.8m/s2 for g, 11kg for m, 99 N for F, and 41° for e. N=(11kg) (9.8m/s) cos 41° +(99N)sin 41° = 146N (b) The net force is, Fret = F cos e-mg sin Rearrange, ma=F cos -mg sino Fcos - mg sino Substitute 9.8m/sfor g, 11kg for m, 99 N for F, and 41° for e. a = (99N) cos 41°-(11kg) (9.8m/s) sin 41° 11kg =0.36m/s2