Question

A 3.25 kg block is sent up a ramp inclined at an angle theta=33.5° from the horizontal. It is given an initial velocity ????0=15.0 m/s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is ????k=0.397 and the coefficient of static friction is ????s=0.712. How far up the ramp in the direction along the ramp does the block go before it comes to a stop?

Answer 1

Answer:

x=13.005m

Explanation:

The kinetic friction μk=0.397 is when the motion is development so:

$F_{k}=u*m*g\\F_{k}=u* m*g cos(33.5)\\F_{k}=0.397*3.25kg*9.8\frac{m}{s^{2}}*cos(33.5)\\F_{k}=10.54 N$

The net force of the motion is the relation of component of block weight acting parallel to ramp and against block's motion

$F_{m}=m*g*sen (33.5) \\F_{m}=3.25kg*9.8\frac{m}{s^{2}}*sen(33.5)\\F_{m}=17.57N$

The net force of the motion is :

$F_{t}=F_{k}+F_{c}\\F_{t}=10.54N+17.57N\\F_{t}=28.11 N$

The total force is the relation of mass and acceleration so, can find the acceleration to determinate the distance the block go far after the motion

$F=m*a\\a=\frac{F}{m}\\a=\frac{28.11N}{3.25kg}\\a=8.65 \frac{m}{s^{2}}$

$v_{f} ^{2}=v_{o} ^{2}+2*a*x\\x=\frac{v_{f} ^{2}-v_{o}^{2}}{2*a}\\x=\frac{0^{2}-15^{2}}{2*8.65}\\x=13.005 m$