Question

1.Calculate the magnitude of acceleration of a 74.2 kg snow boarder going up a 7.52^\circ​∘​​ ​​, slope assuming the coefficient of friction for waxed wood on wet snow (\mu _s = 0.14,~\mu_k = 0.1μ​s​​=0.14, μ​k​​=0.1). Note that not all of these quantities may be needed.

2.The only forces action on a 4.87 kg object are \vec{F}_1​F​⃗​​​1​​ = 1.56\hat{x}​x​^​​ - 8.64\hat{y}​y​^​​ N, \vec{F}_2​F​⃗​​​2​​ = -7.10\hat{x}​x​^​​ + 7.82\hat{y}​y​^​​ N, and \vec{F}_3​F​⃗​​​3​​ = 9.62\hat{x}​x​^​​ + 1.98\hat{y}​y​^​​ N. If the object is initially at rest, how far has it traveled after 2.20 seconds?

3.A sled of mass 2.34 kg has an initial speed of 4.85 m/s across a horizontal surface. The coefficient of kinetic friction between the sled and surface is 0.220. What is the speed of the sled after it has traveled a distance of 3.85 m?

Answer 1

1) net force acting along the snow board, Fnet = -m*g**sin(7.52) - fk

m*a = -m*g*sin(7.52) -mue_k*N

m*a = -m*g*sin(7.52) -mue_k*m*g*cos(7.52)

a = -g*sin(7.52) - mue_k*g*cos(7.52)

= -9.8*sin(7.52) - 0.1*9.8*cos(7.52)

= -2.25 m/s^2

magnitude of acceleration, |a| = 2.25 m/s^2 <<<<<<<------Answer

2)

Fnetx = F1x + F2x + F3x

= 1.56 - 7.1 + 9.62

= 4.08 N

Fnety = F1y + F2y + F3y

= -8.64 + 7.82 + 1.98

= 1.16 N

Fnet = sqrt(Fnetx^2 + Fnety^2)

= sqrt(4.08^2 + 1.16^2)

= 4.24 N

a = Fnet/m

= 4.24/4.87

= 0.871 m/s^2

distance travelled in 2.2 s, d = vo*t + (1/2)*a*t^2

= 0 + (1/2)*0.871*2.2^2

= 2.11 m <<<<<<<<<------------------Answer

3) a = -g*mue_k

= -9.8*0.22

= -2.156 m/s^2

use, v^2 - u^2 = 2*a*d

v = sqrt(u^2 + 2*a*d)

= sqrt(4.85^2 + 2*(-2.156)*3.85)

= 2.63 m/s <<<<<<<<<------------------Answer