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

x=13.005m

Step-by-step 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(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(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=(F)/(m)\\a=(28.11N)/(3.25kg)\\a=8.65 (m)/(s^(2))`

`v_(f) ^(2)=v_(o) ^(2)+2*a*x\\x=(v_(f) ^(2)-v_(o)^(2))/(2*a)\\x=(0^(2)-15^(2))/(2*8.65)\\x=13.005 m`