Question

A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The maximum acceleration the truck can have so the bag does not slip is 2.40 m/s2 . Calculate the (a) angle of the hill relative to horizontal and (b) coefficient of static friction between the bag and the truck. (c) The truck is now travelling on level ground at constant speed. The sand bag is tossed forward sliding along the truck bed with a horizontal speed of 2.55 m/s. If the coefficient of kinetic friction is 0.350, how far does the bag slide before coming to rest

Answer 1

Answer:

a

   $\theta = 23.32^o$

b

  $\mu_s = 0.27$

c

$s = 0.948 \ m$

Explanation:

From the question we are told that

The mass of the bag is $m_b = 25.0 \ kg$

The normal force experienced is $F_n = 225 \ N$

The maximum acceleration of the bag is $a = 2.40 \ m/s^2$

Generally this normal force experience by the bag is mathematically represented as

$F_n = mg cos \theta$

=> $225 = (25 * 9.8) cos \theta$

=> $0.9183 = cos \theta$

=> $\theta = cos^{-1}[0.9183]$

=> $\theta = 23.32^o$

Generally for the bag not to slip , it means that the frictional force is equal to the sliding force

$F_f = F_s$

Hence $F_f$ is mathematically represented as

$F_f = \mu_s * F_n$

While $F_s$ is mathematically represented as

$F_s = m * a$

So

$\mu_s * F_n = m * a$

=> $\mu_s * 225 = 25 * 2.40$

=> $\mu_s = 0.27$

Generally from the workdone equation we have that

$KE_f - KE_i = W_f$

Here $W_f$ is the work done by friction which is mathematically represented as

$W_f = m * g * \mu_k * s$

Here s is the distance covered by the bag

$KE_f$ is zero given that velocity at rest is zero

and

$KE_i = \frac{1}{2} * m* v_i^2$

so

   $\frac{1}{2} * m* v_i^2 = m * g * \mu_k * s$

=>  $\frac{1}{2} * v_i^2 = g * \mu_k * s$

substituting  2.55 m/s for v_i and 0.350 for  \mu_k  we have that

     $\frac{1}{2} * 2.55^2 = 9.8 * 0.350 * s$

=> $s = 0.948 \ m$