If an object with a mass of #10 kg # is moving on a surface at #45 m/s# and slows to a halt after # 6 s#, what is the friction coefficient of the surface?

Answer 1

#u_k=0,0051#

#v_l=v_i-a*t# #v_l:"last velocity"# #v_i:"initial velocity"# #a:"acceleration"# #t:"time"#
#v=v/2-a*6# #6*a=v-v/2# #6*a=-v/2# #a=-v/12" "a=-3,75" " m/s^2" 'acceleration of object'"#
#v_l^2=v_i^2-2*a* Delta x# #v^2=(v/2)^2-2.3,75*Delta x#
#v^2-v^2/4=7,50*Delta x#
#(3*v^2)/4=7,50*Delta x# #3*v^2=30*Delta x# #v^2=10*Delta x" ; "45^2=10*Delta x" ; "2025=10*Delta x# #Delta x=202,5 " m"# #E_k=1/2*m*v^2 "the kinetic energy of object"# #W_f=F_f*Delta x" work doing by friction force"# #F_f=u_k*m*g" "W_f=u_k*m*g*Delta x# #E_k=W_f# #1/2*cancel(m)*v^2=u_k*cancel(m)*g*Delta x#
#u_k=v^2/(2*g*Delta x)#
#u_k=45^2/(2*9,81*202,5)=2025/397305#
#u_k=0,0051#
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Answer 2

We can use the equation of motion to find the friction coefficient of the surface:

[ \text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time}} ]

Given that the object slows to a halt, its final velocity is 0 m/s. So, the change in velocity is ( 45 , \text{m/s} - 0 , \text{m/s} = 45 , \text{m/s} ).

Plugging in the values:

[ \text{Acceleration} = \frac{45 , \text{m/s}}{6 , \text{s}} ]

[ \text{Acceleration} = 7.5 , \text{m/s}^2 ]

Now, we can use Newton's second law to find the force of friction:

[ \text{Force of friction} = \text{mass} \times \text{acceleration} ]

Plugging in the values:

[ \text{Force of friction} = 10 , \text{kg} \times 7.5 , \text{m/s}^2 ]

[ \text{Force of friction} = 75 , \text{N} ]

The force of friction is equal to the force applied to stop the object. Since there are no other forces acting on the object, the force of friction is equal to the force applied:

[ \text{Force of friction} = \text{Force applied} ]

Thus, the friction coefficient can be calculated using the formula:

[ \text{Friction coefficient} = \frac{\text{Force of friction}}{\text{Normal force}} ]

Since the normal force equals the gravitational force, and the mass of the object is ( 10 , \text{kg} ), the normal force is ( (10 , \text{kg}) \times (9.8 , \text{m/s}^2) = 98 , \text{N} ).

Plugging in the values:

[ \text{Friction coefficient} = \frac{75 , \text{N}}{98 , \text{N}} ]

[ \text{Friction coefficient} \approx 0.765 ]

So, the friction coefficient of the surface is approximately ( 0.765 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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