A block of mass M = 3.8 kg is at rest on an incline (θ = 16.0°) that has friction between the incline’s surface and the base of the block. The block sits a distance L from the bottom of the incline. Attached to the block is a stretched spring (i.e., L Xo ) that has a spring constant k = 25.0 N/m and a natural length of Xo = 2.4 meters. Suppose the magnitude of the frictional force acting on the block is Ffr = 19.8 Newtons. Use g = 9.81 m/s2.

Question

A block of mass M = 3.8 kg is at rest on an incline (θ = 16.0°) that has friction between the incline’s surface and the base of the block. The block sits a distance L from the bottom of the incline. Attached to the block is a stretched spring (i.e., L > Xo ) that has a spring constant k = 25.0 N/m and a natural length of Xo = 2.4 meters. Suppose the magnitude of the frictional force acting on the block is Ffr = 19.8 Newtons. Use g = 9.81 m/s2.

anonymous · Answer

1) What is Fs , the magnitude of the spring force acting on the block?

anonymous · Answer

Have you drawn the free body diagram of the block?

anonymous · Answer

Applying Newton's second law, 
$\Sigma F= {F_g} {sin {25}} -F_f -F_s=0 $, static equi. 
Do you need further help?

anonymous · Answer

i have drawn a fbd why sin25

anonymous · Answer

i was trying 19.8-19.8(.7) bcuz i was doin tan16

anonymous · Answer

oh right. sin 16, sorry. |dw:1352866388351:dw|
Taking the forces along the incline. 
Can I see your free body diagram? Why tan16?

anonymous · Answer

so would i do 19.8(sin16)

anonymous · Answer

sorry my fbd is on my ipad its hard for me to draw bcuz i have a spinal cord injury i cant send it

anonymous · Answer

Don't worry about it :)
well, the gravitational force, F_g=3.8*9.8=37.24N
And taking the x-vector along the incline, it's F_g sin 16.