Answer:
a
[tex]\theta = 23.32^o [/tex]
b
[tex] \mu_s = 0.27 [/tex]
c
[tex] s = 0.948 \ m [/tex]
Explanation:
From the question we are told that
The mass of the bag is [tex]m_b = 25.0 \ kg[/tex]
The normal force experienced is [tex]F_n = 225 \ N[/tex]
The maximum acceleration of the bag is [tex]a = 2.40 \ m/s^2[/tex]
Generally this normal force experience by the bag is mathematically represented as
[tex]F_n = mg cos \theta[/tex]
=> [tex]225 = (25 * 9.8) cos \theta[/tex]
=> [tex] 0.9183 = cos \theta[/tex]
=> [tex]\theta = cos^{-1}[0.9183][/tex]
=> [tex]\theta = 23.32^o [/tex]
Generally for the bag not to slip , it means that the frictional force is equal to the sliding force
[tex]F_f = F_s[/tex]
Hence [tex]F_f [/tex] is mathematically represented as
[tex]F_f = \mu_s * F_n [/tex]
While [tex]F_s [/tex] is mathematically represented as
[tex]F_s = m * a [/tex]
So
[tex] \mu_s * F_n = m * a [/tex]
=> [tex] \mu_s * 225 = 25 * 2.40 [/tex]
=> [tex] \mu_s = 0.27 [/tex]
Generally from the workdone equation we have that
[tex]KE_f - KE_i = W_f[/tex]
Here [tex]W_f[/tex] is the work done by friction which is mathematically represented as
[tex]W_f = m * g * \mu_k * s[/tex]
Here s is the distance covered by the bag
[tex]KE_f[/tex] is zero given that velocity at rest is zero
and
[tex]KE_i = \frac{1}{2} * m* v_i^2[/tex]
so
[tex] \frac{1}{2} * m* v_i^2 = m * g * \mu_k * s [/tex]
=> [tex] \frac{1}{2} * v_i^2 = g * \mu_k * s [/tex]
substituting 2.55 m/s for v_i and 0.350 for \mu_k we have that
[tex] \frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s [/tex]
=> [tex] s = 0.948 \ m [/tex]
