Answer:
2.00 m/s²
Explanation:
Given
The Mass of the metal safe, M = 108kg
Pushing force applied by the burglar, F = 534 N
Co-efficient of kinetic friction,
= 0.3
Now,
The force against the kinetic friction is given as:
![f = \mu_k N = u_k Mg](https://tex.z-dn.net/?f=f%20%3D%20%5Cmu_k%20N%20%3D%20u_k%20Mg)
Where,
N = Normal reaction
g= acceleration due to the gravity
Substituting the values in the above equation, we get
![f = 0.3\times108\times9.8](https://tex.z-dn.net/?f=f%20%3D%200.3%5Ctimes108%5Ctimes9.8)
or
![f = 317.52N](https://tex.z-dn.net/?f=f%20%3D%20317.52N)
Now, the net force on to the metal safe is
![F_{Net}= F-f](https://tex.z-dn.net/?f=F_%7BNet%7D%3D%20F-f)
Substituting the values in the equation we get
![F_{Net}= 534N-317.52N](https://tex.z-dn.net/?f=F_%7BNet%7D%3D%20534N-317.52N)
or
![F_{Net}= 216.48](https://tex.z-dn.net/?f=F_%7BNet%7D%3D%20216.48)
also,
acceleration of the safe
Therefore, the acceleration of the metal safe will be
acceleration of the safe=![\frac{F_{Net}}{M}](https://tex.z-dn.net/?f=%20%5Cfrac%7BF_%7BNet%7D%7D%7BM%7D%20)
or
acceleration of the safe=![\frac{216.48}{108}](https://tex.z-dn.net/?f=%20%5Cfrac%7B216.48%7D%7B108%7D%20)
or
acceleration of the safe=![2.00 m/s^2](https://tex.z-dn.net/?f=%202.00%20m%2Fs%5E2%20)
Hence, the acceleration of the metal safe will be 2.00 m/s²