Only part D and E **Problem Statement:**A 3.00-kg object is acted upon by an applied force causing it to move along a rough horizontal plane, with its x and y coordinates given by \( x = 5t^2 - 1 \) and \( y = 3t^3 + 2 \), where \( x \) and \( y \) are in meters and \( t \) is in seconds. The coefficients of friction between the object and the surface are 0.4 and 0.8. Answer the following questions:a. What are the units of the numerical coefficients?b. Determine the acceleration of the object as a function of time.c. Determine the Net force acting on the object at \( t = 5s \).d. Determine the force of friction acting on the object.e. Determine the magnitude and direction of the applied force on the object at \( t = 5s \).

Given data: m=3 kgx=5t2−1y=3t3+2μk=0.4μs=0.8Answer (a) The unit of numerical coefficient of coordinate x can be determined as shown below. x=5t2−1m=ps2−0m=ps2p=m/s2 Also, y=3t3+2m=qs3+0q=m/s3Answer (b) Acceleration along the x axis can be determined as shown below. ax=d2xdt2ax=d25t2−1dt2ax=ddtddt5t2−1ax=ddt10t−0ax=10 m/s2 Acceleration along the y axis can be determined as shown below. ay=d2ydt2ay=d23t3+2dt2ay=ddtddt3t3+2ay=ddt9t2+0ay=18t m/s2The total acceleration of the object can be determined as shown below. a=ax2+ay2a=10 m/s22+18t m/s22a=100+324t2 m/s2Answer (c) The acceleration of the object at time equal to 5 seconds can be determined as shown below. a=100+32452 m/s2a=90.55 m/s2 The net force on the object can be determined as shown below. Fnet=maFnet=3 kg90.55 m/s2Fnet=271.65 NAnswer (d) Since the object is under motion, the kinetic coefficient of friction is taken under consideration. Therefore, the friction force can be determined as shown below. f=μsmgf=0.43 kg9.81 m/s2f=11.772 NAnswer (e) The magnitude of the applied force can be determined as shown below. Fnet=F−f271.65 N=F−11.772 NF=283.42 N The applied force is in the direction of motion of the object.