Question

A 3.25 kg block is sent up a ramp inclined at an angle θ = 32.5 ° from the horizontal. It is given an initial velocity v 0 = 15.0 m / s up the ramp. Between the block and the ramp, the coefficient of kinetic friction is μ k = 0.382 and the coefficient of static friction is μ s = 0.687. How far up the ramp in the direction along the ramp does the block go before it comes to a stop?

Answer 1

Answer:

17.11m

Explanation:

Weight of the block is $mg=3.25\times 9.8=31.85N$, $g=9.8N/kg$

#Component of the weight acting normal to ramp is:

$31.85Ncos \theta=31.85cos 32.5\\=26.862N$

#Component of the weight acting parallel to the ramp and against block's motion :

$31.85Nsin\theta=31.85Nsin 32.5\textdegree\\=17.113N$

Force of kinetic friction parallel to ramp and against block's motion:

$\mu_k=0.382\\=>0.382\times 26.862N=10.2613N$

$F_n_e_t=10.2613+17.113=21.3743N$

We can now calculate deceleration up incline:

$a=F_n_e_t/m\\=21.3743/3.25\\=6.5767m/s^2$

#The time,$t$, of block's motion up incline is given as:

$t=V_O/a\\=15.0/6.5767\\=2.2808s$

$v_a_v_g=(15+0)/2=7.5m/s\\\\d=v_a_v_g(t)=7.5\times2.2808\\=17.1058m$

The block moves 17.11m up the incline.