- #1
gracedescent
- 21
- 0
A boy drags his 60N sled at a constant speed up a 15.0* hill. He does so by pulling with a 25N force on a rope attached to the sled. If the rope is inclined at 35* to the horizontal, (a) what is the coefficient of kinetic friction between sled and snow? (b) At the top of the hill, he jumps on the sled and slides down the hill. What is the magnitude of his acceleration down the slope?
So we know:
(tension of rope) T=25N
(weight of sled) mg=60N
(m of sled) m=6.12kg
(normal force) n=56.36N
(force of gravity) Fg=20.51N
So,
[tex]\sum[/tex]Fx=T*cos[tex]\Theta[/tex]-mg*sin[tex]\Theta[/tex]-fk=max=ma
[tex]\sum[/tex]Fy=T*sin[tex]\Theta[/tex]+n-mg*cos[tex]\Theta[/tex]=may=0?
Then,
[tex]\sum[/tex]Fx=25Ncos20-6.12kg*9.80m/s2*sin20-(uk*56.36N)=6.12kg*ax
ax=25Ncos20-6.12kg*9.80m/s2*sin20-(uk*56.36N)=6.12kg*ax/6.12kg
2.98-(uk*56.36N)/6.12kg=ax
So, since I have two unknowns I am lost now. And I haven't even begun part (b).
Could anyone help me and teach me to find these unknowns?
Thank you!
So we know:
(tension of rope) T=25N
(weight of sled) mg=60N
(m of sled) m=6.12kg
(normal force) n=56.36N
(force of gravity) Fg=20.51N
So,
[tex]\sum[/tex]Fx=T*cos[tex]\Theta[/tex]-mg*sin[tex]\Theta[/tex]-fk=max=ma
[tex]\sum[/tex]Fy=T*sin[tex]\Theta[/tex]+n-mg*cos[tex]\Theta[/tex]=may=0?
Then,
[tex]\sum[/tex]Fx=25Ncos20-6.12kg*9.80m/s2*sin20-(uk*56.36N)=6.12kg*ax
ax=25Ncos20-6.12kg*9.80m/s2*sin20-(uk*56.36N)=6.12kg*ax/6.12kg
2.98-(uk*56.36N)/6.12kg=ax
So, since I have two unknowns I am lost now. And I haven't even begun part (b).
Could anyone help me and teach me to find these unknowns?
Thank you!
Last edited: